hyM.UkliULixiiiULili.itit»iiiiIiiii]yi:iiliih.i 


TREATISE 

OF  A. 

PLANE   TRIGONOMETRY. 

TO    WHICH     IS    PREFIXED 

a  summary  view  of  the  nature  and  use  or 
jLOG.IRITHMS: 

RZING 

THE  SECOND  PART 

OP 

A  COURSE  OF  MATHEMATICS. 

ADArTED  TO  THE  METHOD  OF  INSTRUCTION  TT 
THE  AMERICAN  COLLEGES. 


BY  JEREMIAH  DAY,  D.D.  LL.D 

Prtiidenl  of  Yale  College. 


THE  SECOND  EDITION, 

\V'ITH    ADDITIONS    AND    ALTERATION' 


NEW-HAVEN  : 
PUBLISHED  BY  HOWE  &  SPALDING 

8.  coNVKRSE,  Prin^tr 
1824. 


^%   » ♦ •  '  • 


DISTRICT  OF  CONNECTICUT,   sb. 

Be  it  remembered,  That  on  the  twenty-ninth  day  ot 
^July,  in  the  forty-ninth  year  of  the  Independence  of  the  Uni- 
ted States  of  America,  Jeremiah  Day,  of  the  said  District, 
hath  deposited  in  this  Office  the  title  of  a   Book,  the   right 
whereof  he  claims  as   Author,    in    the   words    following — 
to  wit : 

"  A  treati&e  of  Flanc  Trigonometry ;  to  which  is  prefixed  a  summary 
"view  of  the  nature  and  use  of  I^ogarithms  :  being  the  second  part  of  a 
*'  course  of  Matliematics,  adapted  to  the  method  of  instruction  in  the  Ameri- 
*<  can  Colleges.  By  Jeremiah  Day,  D.  D.  LL.  D.,  President  of  Yale  College. 
"  The  second  edition,  with  additions  and  alterations." 

In  conformity  to  the  Act  of  the  Cono^ress  of  the  United  States,  entitled  "  Au 
Act  for  the  encouragement  of  learning,  by  securing  the  copies  of  Maps, 
Charts,  and  Books,  to  the  Authors  and  Proprietors  of  such  copies,  during  the 
times  therein  mentioned." 

CHAS.  A.  INGERSOLL,  Clerk  of  the  District  of  Connecticut. 
A  true  copy  of  Record,  examined  and  sealed  by  rne, 

CHAS.  A.  INGERSOLL,  Clerk  of  the  District  of  Connecticut. 


r^  Mathemat'ci^I 


nPHE  plan  upon  which  this  work  was  originally  commen- 
-*■  ced,  is  continued  in  this  second  part  of  the  course.  As 
the  single  object  is  to  provide  for  a  class  in  college,  such 
matter  as  is  not  embraced  by  this  design  is  excluded.  The 
mode  of  treating  the  subjects,  for  the  reason*,  mentioned  in 
the  preface  to  Algebra,  is,  in  a  considerable  degree,  diffuse. 
It  was  thought  better  to  err  on  this  extreme,  than  on  the 
other,  especially  in  the  early  part  of  the  course. 

The  section  on  right  angled  triangles  will  probably  be  con- 
sidered as  needlessly  minute.  The  solutions  might,  in  all 
cases,  be  effected  by  the  theorems  which  are  given  for  ob- 
lique angled  triangles.  But  the  applications  of  rectangular 
trigonometry  are  so  numerous,  in  navigation,  surveying,  as- 
tronomy. &c.  that  it  was  deemed  important,  to  render  famil- 
iar the  various  methods  of  stating  the  relations  of  the  sides 
and  angles;  and  especially  to  bring  distinctly  into  view  the 
principle  on  which  most  trigonometrical  calculations  are 
founded,  the  proportion  between  the  parts  of  the  given  tri- 
angle, and  a  similar  one  formed  from  the  sines,  tangents,  fac. 
in  the  tables. 


CONTENTS 


LOGARITHMS. 

Page. 

Section     I.     Nature  of  Logarithms       -----  f 
II.     Directions   for   taking  logarithms   and 

their  numbers  from  the  tables      -     -  10 

III.     Methods  of  calculating   by  logarithms 

Multiplication       .,---.  17 
Division        -.------21 

Involution         -------  22 

Evolution      --------  25 

Proportion       -     -* 27 

Arithmetical  Complement      -     -     -  28 

Compound  Proportion    -     -     -     -  30 

Compound  Interest      -----  32 

Increase  of  Population  -     -     .     -  35 

Exponential  Equations     -     -     -     -  39 

JV.     Different  Systems  of  Logarithms  -     -  42 

Computation  of  Logarithms        -     -  45 

TRIGONOMETRY. 


Section     I.     Sines,  Tangents,  Secants,  &ic.     -     -     -  49 
II.     Explanation    of   the    Trigonometrical 

Tables 58 

III.  Solutions  of  Right  angled  Triangles  -  66 

IV.  Solutions  of  Oblique  angled  Triangles  80 
V.     Geometrical  Construction  of  Triangles  91 

VI.     Description  and  use  of  Gunter's  Scale  97 

Vn.     Trigonometrical  Analysis     -     -     -     -  105 

VIII.     Computation  of  the  Canon     -     -     -  123 

IX.     Particular  Solutions  of  Triangles  -     -  127 

Notes 137 

Table  of  Natural  Sines  and  Tangents  147 


LOGARITHMS. 


SECTION  I 

NATURE  OF  LOGARITHMS.* 


'T'HE  operations  of  Multiplication  and  Division. 
\RT.  1.  J.  ^yjjgj^  ^^Qy  g^pg  ^Q  be  often  repeated,  become 
so  laborious,  that  it  is  an  object  of  importance  to  substitute, 
in  their  stead,  more  simple  methods  of  calculation,  such  as 
Addition  and  Subtraction.  If  these  can  be  made  to  perform, 
in  an  expeditious  manner,  the  otBce  of  multiplication  and 
division,  a  great  portion  of  the  time  and  labour  which  the 
latter  processes  require,  may  be  saved. 

Now  it  has  been  shown,  (Algebra,  233,  237,)  that  powers 
may  be  multiplied,  by  adding  their  exponents,  and  divided, 
by  subtracting  their  exponents.  In  the  same  manner,  roots 
may  be  multiplied  and  divided,  by  adding  and  subtracting 
their  fractional  exponents.  (Alg.  280,  286.)  When  these  ex- 
ponents are  arranged  in  tables,  and  applied  to  the  general 
purposes  of  calculation,  they  are  called  Logarithms. 

2.  LOGARITHMS,  thkn,  are  the  EXPONENTS  of  a 

-CRIES  OF  POWERS  AND  ROOTS. f 

In  forming  a  system  of  logarithms,  some  particular  num- 
ber is  tixed  upon,  as  the  base,  radix^  or  first  power,  whose  log- 
arithm is  always  1.  From  this,  a  series  of  powers  is  raised, 
and  the  exponents  of  these  are  arranged  in  tables  for  use. 
To  explain  this,  let  the  number  which  is  chosen  for  the  first 

*=  Maskal3ne's  Preface  to  Taylor's  Logarithms.  Introduction  to  Ilutton's 
Tahlcs.  Keil  on  Logarithms.  Maseres  Scriptores  Logarithmici.  Briggs' 
Logarithms.     Dodson's  Anti-logarithmic  Canon.     Kuler's  Algfl'nt. 

■J-  See  notn  A. 


S2  NATURE  OF 

power,  be   represented  by  a.     Then  taking  a  series  ol' pow- 
ers, both  direct  and  reciprocal,  as  in  Alg.  207 ; 

a/^f  «^,  «-,  a\  «%  «-',  a~-,  a~^,  a    "•,  &c. 

The  logarithm  of  a^  is  3,  and  the  logarithm  of  a~'  is  —1, 
of  a'  is  1,  of  a~2  ig  _2^ 

offlO  is  0,  of  a-3  is  — 3,&c, 

Universally,  the  logarithm  of  a""  is  x. 

3.  In  the  system  of  logarithms  in  common  use,  called 
Brig^s"^  logarithms,  the  number  which  is  taken  for  the  radix 
or  base  is  10.  The  above  series  then,  by  substituting  10  for 
a.  becomes 

lOS  10^  10%  10%  10%  10-%  10-%  10-%&;c. 
Or  10000,  1000,  100,  10,      1,     ^V,      tU,    toVo,  &ic. 
Whose  logarithms  are 
4,       3,        £!,       1,     0,     —1,       -2,      -3,  &c. 

-1.  The  fractional  exponents  of  roots^  and  of  powers  of 
roots,  are  converted  into  decimals,  before  they  are  inserted 
in  the  logarithmic  tables.     See  Alg.  255. 

The  logarithm  of  a^,  or  a^'^  ^2%  is  0.3333, 
of  a^,  or  «'-^^'^%  is  0.6666, 
of  cr,  or  «^-^-«%  is  0.4285, 
of  «  2'^  or^3.0666js  3.6666,  kc. 

These  decimals  are  carried  to  a  greater  or  less  number  of 
places,  according  to  the  degree  of  accuracy  required. 

5.  In  forming  a  system  of  logarithms,  it  is  necessary  to 
obtain  the  logarithm  of  each  of  the  numbers  in  the  natural 
series  1,  2,  3,  4,  5,  &;c.:  so  that  the  logarithm  of  any  number 
may  be  found  in  the  tables.  For  this  purpose,  the  radix  of 
the  system  must  first  be  determined  upon  ;  and  then  every 
other  number  may  be  considered  as  some  power  or  root  of 
ihis.  If  the  radix  is  10,  as  in  the  common  system,  every 
other  number  is  to  be  considered  as  some  power  of  10. 

That  a  power  or  root  of  10  may  be  found,  which  shall  be 
equal  to  any  other  number  whatever,  or,  at  least,  a  very  near 
approximation  to  it,  is  evident  from  this,  that  the  exponent 
rnay  be  endlessly  varied;  and  if  this  be  increased  or  dimin- 
j-hed.  the  pov:er  will  be  increased  or  diminished. 


LOGARITHM.^.  -^ 

If  tke  exponent  is  a  fraction,  and  the  numerator  be  increas- 
ed, the  power  will  be  increased  :  but  if  the  denominator  be 
increased  the  power  will  be  diminished. 

6.  To  obtain  then  the  logarithm  of  any  number,  accor- 
ding to  Briggs'  system,  we  have  to  tind  a  power  or  root  of  10 
which  shall  be  equal  to  the  proposed  number.  The  expo- 
nent of  that  power  or  root  is  the  logarithm  required.     Thus 

7=100.845  1^  fof       7  is  0.8451 

20=10'-'°»°  I  therefore  the  ,'  of     20  is  1.3010 

30=10^*^' '*  J       logarithm  ]  of     30  is  1.4771 

400=10='^*'2°J  Lof  400  is  2.6020,  &€, 

7.  A  logarithm  generally  consists  of  two  parts,  an  integer 
and  a  decimal.  Thus  the  logarithm  2.60206,  or,  as  it  is  some- 
times written,  2-f  .60206,  consists  of  the  integer  2,  and  the 
decimal  .60206.  The  integral  part  is  called  the  characteris- 
tic or  index*  of  the  logarithm  ;  and  is  frequently  omitted,  in 
the  common  tables,  because  it  can  be  easily  supplied,  when- 
ever the  logarithm  is  to  be  used  in  calculation. 

By  art.  3d,  the  logarithms  of 

10000,  1000,  100,  10,    1,     .1,     .01,  .001,  &c. 
are       4,         3,         2,      1,    0,    -1,   —2,     -3,  &c. 

As  the  logarithms  of  1  and  of  10  are  0  and  1,  it  is  evident, 
that,  if  any  given  number  be  between  1  and  10,  its  logarithm 
will  be  between  0  and  1 ,  that  is,  it  will  be  greater  than  0,  but 
less  than  1.  It  will  therefore  have  0  for  its  index,  with  a 
decimal  annexed. 

Thus  the  logarithm  of  5  is  0.69897. 
For  the  same  reason,  if  the  given  number  be  between 


10  and  100,      ) 

'    the  log. 

Cl   and  2,  i. 

e.  1+the  dec.  part 

100  and  1000,    > 

■     will  be 

<2  and  3, 

2-t-the  dec.  part 

000  and  10000,^ 

,    between 

(3  and  4, 

3-l-the  dec.  part 

We  have,  therefore,  when  the  logarithm  of  an  integer  or 
mixed  number  is  to  be  found,  this  general  rule. 

*  The  term  index,  as  it  is  used  here,  may  possibly  lead  to  some  confusion 
m  the  mmd  of  the  learner.  For  the  logarithm  itself  is  the  index  or  exponent 
of  a  power.     The  characteristic,  therefore,  is  the  index  of  an  index. 


A  NATURE  OF 

8.  Tlic  index  of  the  logarithm  is  ahvays  one  less,  than  the 
number  of  integral  Jigurts,  in  the  natural  number  whose  loga- 
rithm is  sought :  or,  ihe  index  shows  how  far  the  first  figure 
of  the  natural  number  is  removed  from  the  place  of  unit?. 

Thus  the  logarithm  of  37  is  1.5G820. 

Here,  the  number  of  figures  being  two^  the  index  of  the 
logarithm  is  I. 

The  logarithm  of  253  is  2.40312. 

Here,  the  proposed  number  253  consists  oi  three  figures, 
the  first  of  which  is  in  the  second  place  from  the  unit  figure. 
The  index  of  the  logarithm  is  therefore  2. 

The  logarithm  of  62.8  is  1.79796. 

Here  it  is  evident  that  the  mixed  number  62.8  is  between 
10  and  100.  The  index  of  its  logarithm  must,  therefore, 
be  1. 

9.  As  the  logarithm  of  1  is  0,  the  logarithm  of  a  number 
less  than  1,  that  is,  of  any  p rope r/?ac/ion,  must  be  negative. 

Thus  by  art.  3d 

The  logarithm  of  yV  or  .1  is  —1, 
of  T^o  or  .01  is  -  2, 
of  ToVo   or  -001  is  —3,  &c. 


0  0  0 


10.  If  the  proposed  number  is  betzoeen  y|o  and  y^* 
its  logarithm  must  be  between  —2  and  —3.  To  obtain  the 
logjarithm,  therefore,  we  must  either  subtract  a  certain  frac- 
tional part  from  — 2,  or  add  a  fractional  part  to  —  3  ;  that  is, 
we  must  either  annex  a  negative  decimal  to  —2,  or  a  positive 
one  to  —3. 

Thus  ihe  logarithm 

of  .008  is  either  -2 -.09691,  or  — 3  +  .90309.* 

The  latter  is  generally  most  convenient  in  practice,  and  is 
more  commonly  written  3.90309.     The  line  over  the  index 

*  That  these  two  expressions  are  of  the  same  value  will  be  evident,  if  we 
subtract  the  same  quantity,  +.90309  from  each.  The  remainders  will  be 
•qual,  and  therefore  the  quantities  from  which  the  subtraction  is  made  must 
be  equal.       See  note  B. 


LOGARITHMS.  b 

ilcnote?,  that  thai  is  negnlivc,  while  the  decimal  pari   of  the 
logarithm  is  positive. 

of  0.3,       isT47712, 


The  logarithm 


r  oi  u.:5,  IS  1.4/  / 1:;, 
<  of  0.00,  is  ^^77815, 
^  of  0.009,  is  3795424. 


And  universally, 

1 1.  The  negative  index  of  a  log-irithm  shows  hozv  far  the 
first  significant  fg}ire  of  the  natural  numher.  is  removed  from 
the  place  of  units,  on  the  right  ^  in  the  same  manner  asa;?os- 
itive  index  shows  how  far  the  first  figure  of  the  natural  num- 
ber is  removed  from  the  place  of  units,  on  tlie  left.  (Art.  8.) 
Thus  in  the  examples  in  the  last  article, 

The  decimal  3  is  in  the/r5^  place  from  that  of  units. 
6  is  in  the  second  place, 
9  is  in  the  third  place  ; 

And  the  indices  of  the  logarithms  are   1,     2,  and  3. 

12.  It  is  often  more  convenient,  however,  to  make  the  in- 
dex of  the  logarithm  positive,  as  well  as  the  decimal  part. 
This  is  done  by  adding  10  to  the  index. 

Thus,  for  —1,  9  is  written;       for  —2,  8,  &;c. 
Because  —  1-f  10  =  9,  -2+10  =  8,  <Sic. 


AVith  this  alteration, 

^T.90309^  /  9.90309, 

The  logarithm  )Y.90309^  becomes  )  8.90309, 

( '3'.90309 S  1 7.90309,  &:c. 


This  is  making  the  index  of  the  logarithm  10  too  great. 
But  with  proper  caution,  it  will  lead  to  no  error  in  practice. 

13.  The  .^j^mof  the  logarithms  of  two  numbers,  is  the  log- 
arithm of  the  product  of  those  numbers  ;  and  the  difference  of 
the  logarithms  of  two  numbers,  is  the  logarithm  of  the  quotient 
of  one  of  the  numbers  divided  by  the  other.  (Art.  2.)  In 
Briggs'  system,  the  logarithm  of  10  is  1 .  (Art.  3.)  If  therefore 
any  number  be  multiplied  or  divided  by  10,  its  logarithm  will 
be  increased  or  diminished  by  1  :  and  as  this  is  an  integer,  it 
will  only  change  the  index  of  the  logarithm,  without  affecting 
the  decimal  part. 


NATURE  OF 


Thus  the  logarithm  of  4730  is  3.67486, 
And  the  logarithm  of         10  is  1. 


The  loearithm  of  the  product      47300  is  4.6748G 
And  the  logarithm  of  the  quotient     473  is  2.67486 

Here  the  index  ouly  is  altered,  while  the  decimal  part  re- 
mains the  same.      We  have  then  this  important  property, 

14.  The  DECIMAL  PART  of  the  logarithm  of  anij  number  ^^ 
the  same,  as  that  of  the  number  multiplied  or  divided  %  10, 
100,  1000,  he. 

Thus  the  log.  of  45670,  is  4.G5963, 

4567,  3.65963, 

456.7,  2.65963, 

45.67,  1.65963, 

4.567,  a65963, 

.4567,  2:65963,  or  9.65963, 

.04567,  ^65963,  >    8.65963, 
.004567,     3.65963,       7.65963. 

This  property,  which  is  peculiar  to  Briggs'  system,  is  of 
great  use  in  abridging  tlie  logarithmic  tables.  For  when  we 
have  the  logarithm  of  any  number,  we  have  only  to  change 
the  index,  to  obtain  the  logarithni  of  every  other  number, 
whether  integral,  fractional,  or  mixed,  consisting  of  the  same 
significant  iigures.  The  decimal  part  of  the  logarithm  of  a 
fraction  found  in  this  way,  is  always  yoo^/Z/r^.  For  it  is  the 
same  as  the  decimal  part  of  the  logarithm  of  a  whole  nu!n- 
ber. 

15.  in  a  series  effractions  continually  decreasing^  the  neg- 
ative indices  of  the  logarithms  continually  increase.     Thus 

In  the  series  1,       .1,     .01,  .001,  .0001,  .00001,  &£c. 

The  logarithms  are  0,      -1,    -2,    -3,      -4,       —5,     Sic. 

If  the  progression  be  continued,  till  the  fraction  is  reduced 
to  0,  the  negative  logarithm  will  become  greater  than  any  as- 
signable quantity.  The  logarithm  of  0,  therefore,  is  infinite 
and  negative,   (Alg.  447.) 

16.  h  is  evident  also,  that  all  negative  logarithms  belong 
to  fraclions  which  are  between  1  and  0:  \y\i\\e  positive  loija- 


J.OGARITHMS.  'i 

rithms  belong  to  natural  numbers  which  are  greater  than  1. 
A?  the  whole  range  of  nunnberg,  both  positive  and  negative, 
is  thus  exhausted  in  supplying  the  loorarithms  of  integral  and 
fractional  positive  quantities;  there  can  be  no  other  numbers 
to  furnish  logarithms  for  negative  quantities.  On  this  ac- 
count the  logarithm  of  a  negative  quantity  is,  by  some  wri- 
ters, considered  as  impossible.  But  as  (here  is  no  difference 
in  the  multiplication,  division,  involution  &lc.  of  positive  and 
negative  quantities,  except  in  applying  the  signs  ;  they  may 
be  considered  as  all  positive,  while  these  operations  are  per- 
forming by  means  of  logarithms  ;  and  the  proper  signs  may 
be  afterwards  affixed. 

17.  If  a  series  of  numbers  be  in  geometrical  progression^ 
their  logarithms  will  be  in  aiuthmetical  j?rogTes5io/i.  For, 
in  a  geometrical  series  ascending,  the  quantities  increase  by 
a  common  multiplier  ;  (Alg.  436.)  that  is,  each  succeeding 
term  is  the  product  of  the  preceding  term  into  the  ratio. 
But  the  logarithm  of  this  product  is  (he  sum  of  the  logarithms 
of  the  preceding  term  and  the  ratio  ;  that  is,  the  logarithms 
increase  by  a  common  addition,  and  are,  therefore,  in  arith- 
metical progression.  (Alg.  422.)  In  a  geometrical  progres- 
sion descending,  the  terms  decrease  by  a  common  divisor  and 
their  logarithms,  by  a  common  difference. 

Thus  the  numbers  1,  10,  100,  1000,  10000,  &ic.  are  in  ge- 
ometrical progression. 

And  their  logarithms  0,  I,  2,  3,  4,  &£c  are  in  arith- 
metical progression. 

T^niversally,  if  in  any  geometrical  series, 
«  =  the  least  term,  r=the  ratio, 

L  =  its  logarithm,  /=its  logarithm  ; 

Then  the  logarithm  oi  ar    is  L+/,  (Art.  1.) 
of  ar^  is  L-{-2/, 
oi  ar^  is  L-^U,  &c. 

Here,  the  quantities  «,  ar^,  ar^,  ar^,  &LC,  are  iugeo- 
tnetrical  progression.  (Alg.  436.) 

And  their  logarithms  L,  L-\-l,  L-\-^l,  L-i-Sl,  he.  are  in 
arithmetical  progression.  (Alg.  423.) 

*  See  note  C. 


LOGARITHMIC 


THE  LOGARITHMIC  CURVE. 

19.  The  relations  of  logarithms,  and  their  corresponding 
numbers,  may  be  represented  by  the  abscissas  and  ordinates 
of  a  curve.  Let  the  line  AC  (Fig.  1.)  be  taken  for  unity. 
Let  AF  be  divided  into  portions,  each  equal  to  AC,  by  the 
points  1,  2,  3,  &c.  Let  the  line  a  represent  the  radix  of  a 
given  system  of  logarithms,  suppose  it  to  be  1.3  ;  and  let  a^, 
a^,  &c.  correspond,  in  length,  with  the  different  powers  of  a. 
Then  the  distances  from  A  to  1,  2,  3,  &c.  will  represent  the 
logarithms  of  «,  a- ,  a^,  he.  (Art.  2.)  The  line  CH  is  called 
the  logarithmic  curve,  because  its  abscissas  are  proportioned 
to  the  logarithms  of  numbers  represented  by  its  ordinates. 
(Alg.527,) 

20.  As  the  abscissas  are  the  distances  from  AC,  on  the  line 
AF,  it  is  evident,  that  the  abscissa  of  the  point  C  is  0,  which 
is  the  logarithm  of  1=AC.  (Art.  2.)  The  distance  from  A 
to  1  is  the  logarithm  of  the  ordinate  «,  which  is  the  radix  of 
the  system.  For  Briggs'  loijarithms.  this  ought  to  be  ten 
times  AC.  The  distance  from  A  to  2  is  the  logarithm  of  the 
ordinate  a^  ;  from  A  to  3  is  the  logarithm  of  a^,  &ic. 

21.  The  logarithms  of  numbers  less  than  a  unit  are  nega- 
tive. (Art.  9.)  These  may  be  represented  by  portions  of  the 
line  AN,  on  the  opposite  side  of  AC.  (Alg.  507.)  The  or- 
dinates rt~S  ^~^?  «~^i  <^c.  are  less  than  AC,  which  is  taken 
for  unity  -,  and  the  abscissas,  which  are  the  distances  from  A 
to  -1,  -2,  -3,  he.  are  negative. 

22.  If  the  curve  be  continued  ever  so  far,  it  will  never 
meet  the  axis  AN,  For,  as  the  ordinates  are  in  geometrical 
progression  decreasing,  each  is  a  certain  portion  of  the  pre- 
ceding one.  They  will  be  diminished  more  and  more,  the 
farther  they  are  carried,  but  can  never  be  reduced  absolutely 
to  nothing.  The  axis  AN  is,  therefore,  an  asymptote  of  the 
curve.  (Alg.  545.)  As  the  ordinate  decreases,  the  abscissa 
increases;  so  ihat,  when  one  becomes  infinitely  small,  the 
other  becomes  infinitely  great.  This  corresponds  with  what 
has  been  stated,  (Art.  15.)  that  the  logarithm  ofO  hinfinite 
and  nesrative. 


CURVE.  9 

23.  To  find  the  equation  of  this  curve, 

Let  a=the  radix  of  the  system, 
x=any  one  of  the  abscissas, 
y=the  corresponding  ordinate. 

Then,  by  the  nature  of  the  curve,  (Art.  19.)  the  ordinate 
to  any  point,  is  that  power  of  a  whose  exponent  is  equal  to 
the  abscissa  of  the  same  point ;  that  is  (Alg.  528.) 

*  For  other  properties  of  the  logarithmic  curve,  see  Fluxions. 


SECTION  II. 


)lRi:CTIOiNS   FOR  TAKING  LOGARITHiMS  AND 
THEIR  NUMBERS  FROM  THE  TABLES.* 


\  ^'^  1  rP^^E  purpose  which  logarithms  are  intended  to 
ans\yer,  is  to  enable  us  to  perform  arithmetical 
operations  w'lih  greater  expedition,  than  by  the  common  meth- 
ods. Before  anyone  can  avail  himself  of  this  advantage,  he 
must  become  so  familiar  with  the  tables,  that  he  can  readily 
find  the  logarithm  of  any  number  ;  and,  on  the  other  hand, 
ihe  number  to  vvliich  any  logarithm  belongs. 

In  the  common  tables,  the  indices  to  the  logarithms  of  the 
first  100  numbers,  are  inserted.  But,  for  all  other  numbers, 
the  decimal  part  only  of  the  logarithm  is  given  ;  while  the 
index  is  left  to  be  supplied,  according  to  the  principles  in 
arts.  8  and  11. 

25.  To  find  the  logrrithm  of  any  7iumber  betzoeen  1  and 
]  00  •, 

Look  for  the  proposed  number,  on  the  left ;  and  against 
it.  in  the  next  column,  will  be  the  logarithm,  with  its  index. 
Thus 

Tne  log.  of  18  h  1.25527.     The  log.  of  73  is  1.86332. 

96.  To  find  the  loganthm  of  any  number  between  100  and 
1000  ;  or  of  any  number  consisting  of  not  more  than  three 
significant  figures,  with  ciphers  annexed. 

In  the  smaller  tables,  the  three  first  tigures  of  each  num- 
ber, are  generally  placed  in  the  left  hand  column  ;  and  the 
fourth  tigure  is  placed  at  the  head  of  the  other  columns. 

Any  number,  therefore,  between  100  and  1000,  may  be 
found  on  the  left  hand  ;  and  directly  opposite,  in  the  next 
column,  is  the  decimal  part  of  its  logarithm.  To  this  the 
index  must  be  prefixed,  according  to  the  rule  in  art.  8. 


*  The  be?t  En-lish  Tables  are  Hutton's  in  8vo.  and  Taylor's  in  4to.  h 
these,  the  logarithms  are  carried  to  seven  places  of  decimals,  and  proportionsi 
parts  are  placed  in  the  margin.  The  smaller  tables  are  numerous :  ai;  .' 
Avheu  accurately  printed,  aresuHicient  for  common  calculation?. 


THE  LOGARITHMIC  TABLES.  Jl 

The  log.  of  458  is  2.66087,     The  log.  of  935  is  2.97081, 
of  796      2.90091,  of  086       2.58659. 

If  there  arc  ciphers  annexed  to  I'hc  significant  fii^ures,  the 
logarithm  may  be  found  in  a  similar  manner.  For,  by  art. 
14,  the  decimal  p;irt  of  the  logarithm  of  any  number  is  the 
same,  as  that  of  the  number  muhiplied  into  10,  100,  &c.  All 
the  difference  will  he  in  the  index  :  and  this  may  be  supplied 
by  the  same  general  rule. 

The  log.  of  4580  is  3.66087,     The  log.  of  326000  is  5.5 1 322, 
'of  79600      4.90091,  of  8010000     6.90363. 

27.  To  find  the  logarithm  of  any  number  consisting  o/four 
figures^  either  zvithj  or  without,  ciphers  annexed. 

Look  for  the  three  first  figures,  on  the  left  hand,  and  for 
the  fourth  figure,  at  the  head  of  one  of  the  columns.  The 
logarithm  will  bo  found,  opposite  the  three  first  figures,  and 
in  the  column  which,  at  the  head,  is  marked  with  the  fourth 
figure.* 

The  log.  of  6234  is  3.79477,  The  log.  of  783400  is  5.89398, 
of5231      3.71858,  ofG281000     6.79803. 

28.  To  find  the  logarithm  of  a  number  containing  more  than 
FOUR  significant  figures. 

By  turning  to  the  tables,  it  will  be  seen,  that  if  the  dij^er- 
ences  between  several  numbers  be  small,  in  comparison  with 
the  numbers  themselves  ;  the  diflfbrences  of  the  logarithms 
will  be  nearly  proportioned  to  the  differences  of  the  nym- 
hers.     Thus 

The  log.  of  1000  is  3.00000, 

^rmni         Q  nr\r\io  Here   the    di.Tereuces  in   the 

of  1 00 1        3.00043,  ,^^^^j,^,^^  ^^^^  1 ,  2,  3,  4,  &c.  and 

of  1002       3.00087,  the  correspondiDg  diflerences  in 

of  1003       3.00130,  thelo-arithmsare43,   87,    13(', 

of  1004     3.00173,  &c.      ^'^'^''• 

Now  43  is  nearly  half  of  87,  one  third  of  130,  one  fourth 
ofl73,  <Sic. 

Upon  this  principle,  we  may  fiiid  the  logarithm  of  a  num- 
ber v/hich  is  between  two  other  numbers  whose    logarithnift 

*  In  Taylor's,  Ilutton's  and  other  tables, /our  figures  are  placed  in  the  If  ft 
hand  column,  and  Xhejiflh  at  the  to[j  of  the  page. 


IJ  THE  LOGARITHMIC 

are  given  by  the  tables.  Thus  the  logarithm  of  21716  is  not 
to  be  found,  in  those  tables  which  give  the  numbers  to  four 
places  of  figures  only. 

But  by  the  table,  the  log.  of  21  720  is  4.33686 
and  the  log.  of  21 710  is  4.33666 

The  difference  of  the  two  numbers  is  10  ;  and  that  of  the 
logarithms  20. 

Also,  the  difference  between  21710,  and  the  proposed  num- 
ber21716  is  6. 

If,  then,  a  difference  of  10  in  the  numbers 
make  a  difference  of  20  in  the  logarithms  : 

A  difference  of    6  in  the  numbers,  will 
make  a  difference  of  12  in  the  logarithms. 

That  is,  10  :  20:  :6  :  12. 

If,  therefore,  12  be  added  to  4.33666,  the  log.  of  21710  ; 
The  sum  will  be  4.33678,  the  log.  of  21716. 

We  have,  then,  this 

RULE. 

To  find  the  logarithm  of  a  number  consisting  of  more  than 
four  figures  ; 

Take  out  the  logarithm  of  two  numbers,  one  greater,  and 
the  other  less,  than  the  number   proposed  :  Find  the  differ- 
ence of  the  two  numbers,  and  the  difference  of  their   loga- 
rithms :  Take  also  the  difference   between  the  least   of  the 
two  numbers,  and  the  proposed  number.     Then  say. 
As  the  ditrerence  of  the  two  numbers. 
To  the  diffcrciuce  of  their  logarithms  ; 
So  is  the  difference  between  the  least  of  the  two 

numbers,  and  the  proposed  number. 
To  the  proportional  part  to  be  added  to 
the  least  of  the  two  logarithms. 
It  will  generally  be  expedient  to  make  ihafour  first  figures^ 
in  the  least  of  the  two  numbers,  ihe  same  as  in  the  proposed 
number,  subi-tituting  ciphers,  for  the  remaining  figures  ;  and 
to  make  the  greater  number  the  same  as  the   less,  with  the 
addition  of  a  unit  to  Hie  last  significant  fif^ure.     Thus 

For  36843,  take  36840,  and  36850, 
For  792674,         792600,        792700, 
For  6537825,      6537000.     6538000,  kc. 


TABLES.  1.; 

The  first  term  of  the  proportion  will  then  he    10,  or  100, 
or  1000,  &c. 

Ex.  1 .     Required  the  logarithm  of  362572. 

The  logarithm  of  362600  is  5.55043 
of  362500      555931 


The  differences  are      100,     and     12. 

Then   100  :  12::  72  :  8.64,  or  9  nearly. 
And  the  log.  o,55d3\  +9  =  5.55940,  the  log.  required. 

Ex.  2.      The  log.  of  78264        is  4.89356 

3.  The  log.  of  143542     is  5.15698 

4.  The  log.  of  1 129535  is  6.05290. 

By  a  little  practice,  such  a  facility,  in  abridging  these  cal- 
culations, may  be  acquired,  that  the  logarithms  may  he  taken 
out,  in  a  very  short  time.  When  great  accuracy  is  not  re- 
quired,it  will  he  easy  to  make  an  allowance  sufficiently  near, 
without  formally  stating  a  proportion.  In  the  larger  tables, 
the  proportional  parts  which  are  to  be  added  to  the  loga- 
rithms, are  already  prepared,  and  placed  in  the  margin. 

29.   Tofnd  the  logarithm  of  a  decimal  fraction. 

The  logarithm  of  a  decimal  is  the  same  as  that  of  a  whole 
number,  excepting  the  index.  (Art.  14.)  To  find  then  the 
logarithm  of  a  decimal,  take  out  that  of  a  whole  number 
consisting  of  the  same  figures  ;  observing  to  make  the  negative 
index  equal  to  the  distance  oj  the  first  sie;nificant  figure  ofthf 
fractionfrom  the  place  oi  units,  (Art.  1 1.) 

The  log.  of  0.07643,       is  2788326,  or  8.88326,  (Art.  12.) 
of  0.00259,  3.41330,  or  7.41330, 

of  0.0006278,      7.79782,  or  6.79732. 

.30.    To  find  the  logarithm  of  a  mixed  decimal  number. 
Find  the  logarithm,  in  the  same  manner  as  if  «//   the  fig- 
ures were  integers  ;  and  then  prefix  the  index  which  belongs 
to  the  integral  part,  according  to  art.  8. 

The  logarithm  of  26.34  is  1.42062. 

The  index  here  is  1,  because  1    is  the  index   of  the  loga 
rithm  of  every  number  greater   than  10,  and  less  than  100. 
(Art.  7.) 


14  THE  LOGARITHiMlC 

The  log.  of  2.36  is  0.37291,  The  log.  of  364.2  is  2.56134, 
of  27.8      1.44404,  of  69.42      1.84148. 

31 .   To  find  the  logarithm  of  a  vulgar  fraction. 

From  the  nature  of  a  vulgar  fraction,  the  numerator  may 
be  considered  as  a  dividend,  and  the  denominator  as  a  divisor  ; 
in  other  words,  the  value  of  the  fraction  is  equal  to  the  quo- 
tient, of  the  numerator  divided  by  the  denominator.  (Alg. 
135.)  But  in  logarithms,  division  is  performed  by  subtrac- 
tion ;  that  is,  the  difference  of  the  logarithms  of  iwo  num- 
bers, is  tho  logarithm  of  the  quotient  of  those  numbers.  (Art. 
1.)  To  find  then  the  logarithm  of  a  vulgar  fraction,  subtract 
the  logarithm  of  the  denominator  from  that  of  the  numerator. 
The  difference  w^ill  be  the  logarithm  of  the  fraction.  Or  the 
logarithm  may  be  found,  by  first  reducing  the  vulgar  fraction 
to  a  decimal.  If  the  numerator  is  less  than  the  denominator, 
the  index  of  the  logarithm  must  be  negative,  because  the  val- 
ue of  the  fraction  is  less  than  a  unit.  (Art.  9.) 

Required  the  logarithm  of  |^. 

The  log.  of  the  numerator  is  1.53148 
of  the  denominator  1.93952 


of  the  fraction        K59196,  or  9.59196. 

The  logarithm  of  ^VA  >s   ^66362,  or  8.66362. 

of_JL-        3.04376,  or  7.04376. 

32.   If  the  logarithm  of  a  mixed  number  \s  required,  reduc* 
it  to  an  improper  fraction,  and  then  proceed  as  before. 

The  logarithm  of  3^=- V  is  0.  57724. 


33.  To  find  the  natural  number  belonging  to  an-^  loga- 
rithm. 

In  computing  by  logarithms,  it  is  necessary,  in  the  first 
place,  to  take  from  the  tables  the  logarithms  of  the  numbers 
which  enlcrinto  the  calculation  ;  and,  on  the  other  hand,  al: 
the  close  of  the  operation,  to  find  the   number  belonging  to 


rABLES.  i., 

the  loganihm  obtained  in  the  result.     This  is  evidently  done, 
hy  reversing  the  methods  in  the  preceding  articles. 

Where  great  accuracy  is  not  required,  look  in  the  tables  for 
the  logarithm  which  is  nearest  to  the  given  one  ;  and  directly 
opposite,  on  the  left  hand,  will  be  found  the  three  first  fig- 
ures, and  at  the  top,  over  the  logarithm,  the  fourth  figure,  of 
the  number  required.  This  number,  by  pointing  off  decim- 
als, or  by  adding  ciphers,  if  necessary,  must  be  made  to  cor- 
respond with  the  index  of  the  given  logarithm,  according  to 
arts.  8  and  11. 

The  natural  number  belonging 

to  3.86493  is  7327,  to  1.62672  is  42.24, 

to2.90141_     796.9,         to^89115       0.07783. 

in  the  last  example,  the  index  requires  that  the  first  signifi- 
cant figure  should  be  in  the  second  place  from  units,  and 
therefore  a  cipher  must  be  prefixed.  In  other  instances,  it  is 
necessary  to  anjiex  ciphers  on  the  right,  so  as  to  make  the 
number  of  figures  exceed  the  index  by  1. 

The  natural  number  belonirinti 

to  6.71 5i3-Z  is  5196000,     to   T.6.5677  is  0.004537, 
to  4.67062  46840,     to  "iTo 9 302     0.0003963. 

34.  When  great  accuracy  is  required,  and  the  given  loga- 
rithm is  not  exactly,  or  very  nearly,  found  in  the  tables,  it 
will  be  necessary  to  reverse  the  rule  in  art.  28. 

Take  from  the  tables  two  logarithms,  one  the  next  greater, 
the  other  the  next  less  than  the  given  logarithm.  Find  the 
difference  of  the  two  logarithms,  and  the  difference  of  their 
natural  numbers;  also  the  difference  between  the  least  of  the 
two  logarithms,  and  the  given  logarithm.     Then  say. 

As  the  difference  of  the  two  logarithms, 
To  the  difference  of  their  numbers  ; 
So  is  the  difference  between  the  given 

logarithm  and  the  least  of  the  other  two. 
To  the  proportional  part  to  be  added  to 

the  least  of  the  two  numbers. 


16  THE  LOGARITHMIC  TABLES. 

Required  the  number  belonging  to  the  logarithm  2.67325. 

Nextgreatlog.  2.67330.  Its  numb.  471.3.  Given  log.  2.67325. 
Next  less  2.67321.  Its  numb.  471.2.  Next  less    2.67321. 


Differences  9  0.1 


Then  9  :  0  .1  !  :4  :  0.044,  which   is    to  be  added 
to  the  number   471.2 


The  number  required  is  471.244. 

The  natural  number  belonging 

to  4.37627  is  23783.45,  to  J_.73698  is  54.57357, 

to  3.69479      4952.08,  toT7o9214is   0.123635. 


35.  Correction  of  the  tables.  The  tables  of  logarithm> 
have  been  so  carefully  and  so  repeatedly  calculated,  by  the 
ablest  computers,  that  there  is  no  room  left  to  question  their 
general  correctness.  They  are  not,  however,  exempt  from 
the  common  imperfections  of  the  press.  But  an  errour  of 
this  kind  is  easily  corrected,  by  comparing  the  logarithm 
with  any  two  others  to  whose  sum  or  difference  it  ought  to  be 
equal.   (Art.  1.) 

Thus  48=24X2  =  16X3  =  12X4=:8X6.  Therefore,  the 
logarithm  of  48  is  equal  to  the  sum  of  the  logarithms  of  24 
and  2,  of  16  and  3,  &ic. 

And3  =  |=:V  =  ¥  =  '/  =  V>  <^c.  Therefore,  the  loga- 
rithm of  3  is  equal  to  the  difference  of  the  logarithms  of  6  and 
2,  of  12  and  4,  &c. 


SECTION  III. 

METHODS  OF  CALCULATING  BY  LOGARITHMS. 


.  -    ^PHE   arithmetical    operations   for  which  loga- 

^^*      *  rithms  were  originally  contrived,  and  on  which 

theirgreat  utility  depends,  are  chiefly  multiplication,  division, 
involution,  evolution,  and  finding  the  term  required  in  single 
and  compound  proportion.  The  principle  on  which  all  these 
calculations  are  conducted,  is  this; 

If  the  logarithms  of  two  numbers  be  added,  the  sum  will  be 
the  logarithm  of  the  product  of  the  numbers  ;  and 

If  the  logarithm  of  one  number  be  subtracted  from  that  of  an- 
other, the  iJiFFERENCE  will  be  the  logarithm  of  the  quotient 
of  one  of  the  numbers  divided  by  the  other. 

In  proof  of  this,  we  have  only  to  call  to  mind,  that  loga- 
rithms are  the  exponents  of  a  series  of  powers  and  roots. 
(Arts.  2,  5.)  And  it  has  been  shown,  that  powers  and  roots 
are  multiplied  by  adding  their  exponents;  and  divided,  by 
subtracting  their  exponents.  (Alg.  233,  237,  280,  286.) 


MULTIPLICATION  BY  LOGARITHMS. 
37.  ADD  the  logarithms  of  the  FACTORS  :  the  SUM 

WILL   BE   THE    LOGARITHM  OF   THE  PRODUCT. 

In  making  the  addition,  1   is  to  be   carried,  for  every  10, 
from  the  decimal  part  of  the  logarithm,  to  the  index.  (Art.  7.) 

Numbers.  Logarithms,  Numbers.     Logarithms, 

Mult,  36.2  (Art.  30.)  1.55871.        Mult.  640         2.80618 
Into     7.84  0.89432.        Into  2.316         0.36474 


Prod.    283.8  2.45305       Prod.  1482         3.17092 


The  logarithms  of  the  two  factors  are  taken  from  the  ta- 
bles. The  product  is  obtained,  by  finding,  in  the  tables,  the 
natural  number  belonging  to  the  sum.  (Art.  33.) 

4 


18  MULTIPLICATION  BY 

Mult.  89.24  1.95056  Mult.  134.  2.12710 

Into     3.687         0.56667  Into     25.6  1.40824 


Prod.  329.  2.51723  Prod.  3430  3.53534 


38.  When  any  or  all  of  the  indices  of  the  logarithms  are 
negative,  they  are  to  be  added  according  to  the  rules  for  the 
addition  of  positive  and  negative  quantities  in  algebra.  But 
it  must  be  kept  in  mind,  that  the  decimal  part  of  the  loga- 
rithm is  ;)05?7tre.  (Art.  10.)  Therefore,  that  which  is  car- 
ried from  the  decimal  part  to  the  index,  must  be  considered 
positive  also. 

Mult.  62.84  1.79824  Mult.  0.0294  2^46835 

Into     0.682        T.83378  Into     0.8372  1.92283 


Prod.  42  86  1.63202  Prod.  0.0246  2.39118 


In  each  of  these  examples,  +1  is  to  be  tarried  from  the 
decimal  part  of  the  logarithm.  This  added  to  —  1,  the  lower 
index,  makes  it  0 ;  so  that  there  is  nothing  to  be  added  to 
the  upper  index. 

If  any  perplexity  is  occasioned,  by  the  addition  of  positive 
and  negative  quantities,  it  may  be  avoided,  by  borrowing  10 
to  the  index.   (Art.  12.) 


Mult.  62.84 
Into      0.682 

1.-9824 
9.83378 

Mult. 
Into 

Prod. 

0.0294 
0.8372 

8.46835 
9.92283 

Prod.   42.86 

1.63202 

0.0246 

8.39118 

Here   10  is  added  to  the  negative  indices,  and  afterwards 
rejected  from  the  index  of  the  sum  of  the  logarithms. 

Multiply       2G.83        J^.42862  1.42862 

Into  0.00069  4.83885  or    6.83885 


Product    0.0185  2.26747  8.26741 


LOGARITHMS.  10 

Here  -f-l  carried  to  —4  makes  it  —  3,  which  added  to  the 
upper  index  +1,  gives  —2  for  the  index  of  the  sum. 


Multiply     .00845  3.92686    or    7.92686 

Into  1068.  3.02857  3  02857 


Product     9.0246  0.95543  0.95543 


The  product  of  0.0362     into  25.38       is  0.9188 
of  0.00467  into  348.1       is  1.625 
of  0.0861      into  0.00843  is  0.0007258 

39.  /7«y  number  of  factors  may  be  multiplied  together  by 
adding  their  logarithms.  If  there  are  several  positive^  and 
several  negative  indices,  these  are  to  be  reduced  to  one,  as 
in  algebra,  by  takiiig  the  difference  between  the  sum  of  those 
which  are  negative,  and  the  sum  of  those  which  are  positive, 
increased  by  what  is  carried  from  the  decimal  part  of  the 
logarithms.  (Alg.  78.) 

Multiply     6832  3.83455  3.83455 

Into         0.00C63  X^93601  or  7.93601 

And         0.651  _K81358  9.81358 

And         0.0231  "2.36361  or   8.36361 

And         62.87  1.79344  1.79844 


Prod.     55.74  1.74619  ^.74619 


Ex.   2.   The  prod,  of  36.4X7.82X68.91X0.3846  is  7544] 

3.  The  prod,  of  0.00629  X2. 647  X0.082  X  278.8  X0.00063 
is  0.0002398. 

40.  Negative  quantities  are  multiplied,  by  means  of  loga- 
rithms, in  the  same  manner  as  those  which  are  positive.  (Art. 
16.)  But,  after  the  operation  is  ended,  the  proper  sign  must 
be  applied  to  the  natural  number  expressing  the  product,  ac- 
cording to  the  rules  for  the  multiplication  of  positive  and 
negative  quantities  in  algebra.     The  negative  index  o(di  log- 


20  MULTIPLICATION  BY 

arithm,  must  not  be  confounded  with  the  sign  which  denotes 
that  the  natural  number  is  negative.  Th:it  which  the  index 
of  the  logarithm  is  intended  to  show,  is  not  whether  the  nat- 
ural number  is  positive  or  negative^  but  whether  it  is  greater 
or  less  than  a  imit.  (Art.  16.) 

Mult,   -f-36.42        1.56134         Mult.  —2.681        0.42330 
Into     —67.31        1.82808         Into     +37.24        1.57101 


Prod. —2451        3.38942         Prod. —99.84        1.99931 


In  these  examples,  the  logarithms  are  taken  from  the  ta- 
bles, and  added,  in  the  same  manner,  as  if  both  factors  were 
positive.  But  after  the  product  is  found,  the  negative  sign 
is  prefixed  to  it,  because  +  is  multiplied  into — .  (Alg.  105.) 

Mult.  0.263  j741996        Mult.  0.065        "^81291 

Into    0.00894        T.95134         Into  0.693         T84073 


Prod.  0.002351       3.37130     Prod.  0.04^)04         2.5364 


Here,  the  indices  of  the  logarithms  are  negative,   but   the 
product  is  positive,  because  the  factors  are  both  positive. 

Mult. —62.59  1.79650         Mult. —68.3  183442 

Into    —0.00863    "3'.93601  Into     --0.0096     "3.98227 


Prod.  -1-0.5402        1.73251  Prod. -f  0.6557       1.81669 


LOGARITHMi^.  21 


Division  nv  Logarithms, 


41.  From  THE  logarithm  of  the  DIVIDEND,  SUB- 
TRACT THE  LOGARITHM  OF  THE  DIN  ISOR  ;  THE  DIF- 
FERENCE WILL  BE  THE  LOGARITHM  OF  THE  QUOTIENT. 

(Art.  36.) 

Numbers.     Logarithms.  Numbers.     Logarithms. 

Divide  6238         3.79505  Divide      896.3         2.y5245 

Bv         2982         3.47451  By  9.847         0.89330 


Quot.     2.092  0.32054  Quot.       91.02  1.959 


42.  The  decimal  part  of  the  logarithm  may  be  subtracted 
as  in  common  arithmetic.  But  for  the  indices,  when  either  of 
them  is  negative,  or  the  lower  one  is  greater  than  the  upper 
one,  it  will  be  necessary  to  make  use  of  the  general  rule  for 
subtraction  in  algebra  ;  that  is,  to  chanjie  the  signs  of  the 
subtrahend,  and  then  proceed  a- in  addition.  (Alg.  82,)  When 
1  is  earned  i'rom  the  decimil  part,  this  is  to  be  considered 
affirmative,  and  applied  to  the  index,  before  the  sign  is 
changed. 

Divide  0.8697         7793937     or     9.93937 
By  98.65  1.99410  1.99410 


Quot.  0.008816        3.94527  7.94527 


In  this  example,  the  upper  logarithm  being  less  than  the 
lower  one,  it  is  necessary  to  borrow  10,  as  in  other  cases  of 
subtraction  ;  and  therefore  to  carry  1  to  the  lower  index, 
which  then  becomes  +2.  This  changed  to  — 2,  and  added 
to  — 1  above  it,  makes  the  index  of  the  difference  of  the  log- 
arithms — 3. 

Divide  29.76  1.47.363  1.47363 

By  6254  3.79616  3.7Q616 


Quot.  0.00476  3.67747   or    7.67747 


22  INVOLUTION  BY 

Here,  1  carried  to  the  lower  index,  makes  it  -{-4.  This 
changed  to  — 4,  and  added  to  1  above  it,  gives  — 3  for  the 
index  of  the  difference  of  the  logarithms. 

Divide  6.832         0-83455         Divide  0.00634  3.80209 

By         .0362         2.55871  By        62.18  1.79365 

Quot.  188.73       2.27584  Quot.  0.000102  4^0844 


The  quotient  of  0.0985  divided  by  0.007241,  is  13.6. 
The  quotient  of  0.0621  divided  by  3.68,  is  0.01687. 

43.  To  divide  negative  qnnntities,  proceed  in  the  same 
manner  as  if  they  were  positive,  (Art.  40  )  and  prefix  to  the 
quotient,  the  sign  which  is  required  by  the  rules  for  division 
in  algebra. 

Divide  +3642  3.56134  Divide —0.657  r81757 

By         —23.68        1.37438  By  -f0.0793         2789927 

Quot.  —153.8        2.18696  Quot.  —0.285  0.91830 

In  these  examples,  the  sign  of  the  divisor  being  different 
from  that  of  the  dividend,  the  sign  of  the  quotient  must  be 
negative.  (Alg.  123.) 

Divide— 0.364        T.56110         Divide —68  5  ^83569 

By         —2.56  0.40824  By  -f  0.094        Y.913]3 

Q„ot.    4-0.1422       T 15286  Quot.  —728.7  2.86256 


Involution  by  Logarithms. 

44.   Involving  a  quantity  is  multiplying  it   into  itself.     By 
means  of  logarithms,  multiplication  is  performed  by  addition 
If,  thon.  the  logarithm  of  any   quantity  be  added  to  itself,  the 


LOGARITHMS.  23 

lofjanthm  o{  a  power  of  that  quantity  will  be  obtained.  But 
adding  a  logarithm,  or  any  other  quantity,  to  itself,  is  multi- 
plication. The  involution  of  quantities,  by  means  of  loga- 
rithms, is  therefore  perfornied,  by  multiplying  the  logarithms. 

Thus  the  logarithm 

of  100  is  2 

of  100X100,     that  is,  of  1002  is  2+2  =2X2. 

of  100X100X100,  2^3  is  24-2  +  2  =2X3. 

of  100X100X100X100,  100*  is  2  +  2  +  2  +  2     =2X4. 

On  the  same  principle,  the  logarithm  of  lOO"  is  2Xn. 
And  the  logarithm  of  x" ,  is  (log.  x)Xn.     Hence, 

45.  To  involve  a  quantity  by  logarithms.     MULTIPLY 

THE  LOGARITHM  OF  THE  QUANTITY,  BY  THE  INDEX  OF  THE 
I'OWER   REQUIRED. 

The  reason  of  the  rule  is  also  evident,  from  the  considera- 
tion, that  logarithms  are  the  exponents  of  powers  and  roots, 
and  a  power  or  root  is  involved,  by  multiplying  its  index 
into  the  index  of  the  power  required.  (Alg.  220,288.) 

Ex.   1.  What  is  the  cube  of    6.296? 
Root      6.296,         its  log.    0.79906 
Index  of  the  power  3 

Power  249.6  2  39718 


2.   Required  the  4th  power  of  21.32 

Root  21.32  log.         1.32879 

Index  4 


Power  206614  5.31516 


3.  Required  the  6th  power  of  1.689 
Root     1.689  log.     0,2276 


Index  G 


Power  23.215  1.36578 


I 


24 


INVOLUTION  BY 


4.  Required  the  I44th  power  of  1.003 

Root     1.003  log.     0.00130 

Index  144 


Power  1.539 


0.18720 


46.  It  must  be  observed,  as  in  the  case  of  multiplication, 
(Art.  38.)  that  what  is  carried  from  the  decimal  part  of  the 
logarithm  \s  positive,  whether  the  index  itself  is  positive  or 
negative.  Or,  if  10  be  added  to  a  negative  index,  to  render 
it  positive,  (Art.  12.)  this  will  be  multiplied,  as  well  as  the 
other  figures,  so  that  the  logarithm  of  the  square,  will  be  20 
too  great  ;  of  the  cube,  30  too  great,  he. 


Ex.   1.  Required  the  cube  of  0.0649 
Root     0.0649  log.  "2.81224 


Index 


Power  0.0002733 


4.43672 


or  8.81224 
3 

6.43672 


2.  Required  the  4th  power_of  0.1234 
Root     0.1234  log.     1.09132  or  9.09132 

Index  4  4 


Power  0.0002319 


4.36528 


6.36528 


3.   Required  the  6th  power  of  0.9977 

Root     0.9977         ]o<r,     1.99900  or  9.99900 


Index 


Power  0.9863 


1.99400 


9.99400 


4.  Required  the  cube  of  0.08762. 
Root     0.08762         log72.94260         or  8.94260 
Index  3  3 


Power  0.0006727 


4.82780 


6.82780 


LOGARITHMS. 

d.  The  7th  power  of  0.9061  is  0.3015. 
6.  The  5th  power  of  0.9344  is  0.7123. 


Evolution  by  Logarithms. 

47.  Evolution  is  the  opposite  of  involution.  Therefore 
as  quantities  are  involved,  by  the  multiplication  of  logarithms, 
roots  are  extracted  by  the  division  of  logarithms  ;  that  is, 

To  extract  the   root  of  a  quantity  by  logarithms,  DIVIDE 

1  HE  LOGARITHM   OF  TflE   QUANTITY,  BY  THE  NUMBER  EXPRES- 
SING THE   ROOT   REQUIRED. 

The  reason  of  the  rule  is  evident  also,  from  the  fact,  that 
logarithms  are  the  exponents  of  powers  and  roots,  and  evo- 
lution is  performed,  by  dividing  the  exponent,  by  the  number 
expressing  the  root  required.  (Alg.  257.) 

1.  Required  the  square  root  of  648.3. 

Numbers.  Logarithms. 

Power  648.3  2)2.81178 

Root      25.46  1.40589 

2.  Required  the  cube  root  of  897.1- 

Power  897.1  3)2.95284 

Root     9.645  0.98428 

In  the  first  of  these  examples,  the  logarithm  of  the  given 
number  is  divided  by  2  ;  in  the  other,  by  3. 

3.  Required  the  10th  root  of  6948. 

Power  6948  10)3.84186 

Root  2.422  0.38418 

4.  Required  the  100th  root  of  983. 

Power  983  100)2.99255 

Root   1.071  0.02992 

The  division  is  performed  here,  as  in  other  cases  of  decim- 
als, by  removing  the  decimal  point  to  the  left. 


26  EVOLUTION  BY 

b.   What  is  the  ten  thousandth  root  of  49680000  ? 

Power  49G80000  10000)7.69618 

Root     K00179  0.00077 

We  have,  here,  an  example  of  the  great  rapidity  with  which 
arithmetical  operations  are  performed  by  logarithms. 

43.  If  the  index  of  the  logarithm  is  negative,  and  is  not  di- 
visible by  the  given  divisor,  without  a  remainder,  a  difficulty 
will  occur,  unless  the  index  be  altered. 

Suppose  the  cube  root  of  0.0000892  is  required.  The 
logarithm  of  this  is  5.95036.  If  we  divide  the  index  by  3, 
the  quotient  will  be — 1,  with — 2  remainder.  This  remain- 
der, if  it  were  positive,  might,  as  in  other  cases  of  division, 
be  prefixed  to  the  next  figure.  But  the  remainder  is  nega- 
tive.^whWe  the  decimal  part  of  the  logarithm  is  positive  ;  so 
that,  when  the  former  is  prefixed  to  the  latter,  it  will  make 
neither  4-2.9  nor — 2.9,  but — 2-f -9.  This  embarrassing  in- 
termixture of  positives  and  negatives  may  be  avoided,  by 
adding  to  the  index  another  negative  number,  to  make  it  ex- 
actly divisible  by  the  divisor.  Thus,  if  to  the  index  — 5  there 
be  added — 1,  the  sum — 6  will  be  divisible  by  3.  But  this 
addition  of  a  negative  number  must  be  compensated,  by  the 
addition  of  an  equal  positive  number,  which  may  be  prefixed 
to  the  decimal  partol  the  logarithm.  The  division  may  then 
be  continued,  without  difficulty,  through  the  whole. 

Thus,  if  the  logarithm ^95036  be  akered  io~6-\-\.d503G 
it  may  be  divided  by  3,  and  the  quotient  will  be  2^5012. 
We  have  then  this  rule, 

49.  Add  to  the  index ,  if  necessary,  such  a  negative  number 
as  will  make  it  exactly  divisible  by  the  divisor,  and  prefix  an 
equal  positive  number  to  the  decimal  part  of  the  logarithm* 

1.  Required  the  5th  root  of  O.Oq2642 

Power  0.009642  lo^^     3.98417 

or  5  +  2^8417 

Root  0.3952  lT59683 

2.  Required  the  7th  root  of  0  0004935 

Power  0.0004935  log.   4.69329 

or  7y7-f3.69^29 

Root  0.337      *        1.52761 


LOGARITHMS  27 

50.  If,  forthe  sake  of  performing  the  division  convenientlj, 
the  negative  index  be  rendered  popitivc.  it  will  be  expedient 
to  borrow  as  manv  tens,  as  there  are  units  in  die  number  de- 
noting the  root. 

What  is  the  fourth  root  of  0.03698  ? 

Power   0.03698  4)^56797  or  4)3^.56797 

Root       0.4385  T.64199  9.64199 

Here  the  index,  by  borrowing,  is  made  40  too  great,  that 
is, +  38  instead  of — 2.  When,  therefore,  it  is  divided  bv  4, 
it  is  still  10  too  great,  -f-9  instead  of  — 1. 

What  is  the  5th  root  of  0.008926  ? 

Power  0.008926  5)3.95066  or  5)47.95066 

Root     0.38916  1.59013  9.50013 

51.  A  power  of  a  roo^  may  be  found  by  first  multip!f/i>icr 
the  logarithm  of  the  given  quantity  into  the  index  of  the 
power,  (Art.  45.)  and  then  dividing  the  product  by  the  num- 
ber expressing  the  root.   (Art.  47.) 

1.  What  is  the  value  of  (53)'',  that  is,  the  6th  power  of 
the  7th  root  of  53  ? 

Given  number  53         log.  1.7242S 
Multiplying  l)y  6 

Dividing  by        7)10.34568 
Power  required  30.06     1.47795 

2.  What  is  the  8th  power  of  the  9th  root  of  654  .' 


Proportio.v  by    Logarithms. 

52.  In  a  proportion,  when  three  terms  are  given,  tlie  fourth 
is  found,  in  common  arithmetic,  by  multiplying  together  the 
second  and  third,  and  dividing  by  the  first.  But  when  lo^a- 
arithms  are  used,  addition  takes  the  place  o  "multiplication, 
and  subtraction,  of  division. 

To  find  then,  by  locrarithms,  the  fourth  term  in  a  propor- 
tion, ADD  THK  LOGARITHMS  OF  THE  SECOND  AND  THIRD 

TERMS.  AND  from  the  .^7ir/*  SUBTRACT  the  LOGARiTHnr  of 


28  ARITHMETICAL 

THE  FIRST  TER3I.     Tlic  remainder  will  be  the  logarithm  oi 
the  term  required. 

Ex.  1.  Find  a  fourth  proportional  to  7964,  378,  and  27960. 

Numbers.  Logarithms. 

Second  term       378  2.57749 

Third  term     27960  4.44654 


7.02403 
First  term         7964  3.90113 


Fourth  term     1327  3.12290 

2.  Find  a  4th  proportional  to  768,  381  and  9780. 
Second  term     381  2.58092 

Third  term     9780  3.99034 


6.57126 
First  term        7G8  2.88536 


Fourth  term  4852  3.68590 


Arithmetical  Complement. 

53.  When  one  number  is  to  be  subtracted  from  another, 
it  is  often  convenient,  first  to  subtract  it  from  10,  then  to  add 
the  difference  to  the  other  number,  and  afterwards  to  reject 
the  10. 

Thus,  instead  o( a — 6,  we  may  put  10  — h-\-a — 10. 

In  the  first  of  these  expressions,  b  is  subtracted  from  a.  In 
the  other,  b  is  subtracted  from  10,  the  difference  is  added  to 
f/,  and  10  is  afterwards  taken  from  the  sum.  The  two  ex- 
pressions arc  equivalent,  because  they  consist  of  the  same 
terms,  with  the  addition,  in  one  of  them,  of  10— 10=0.  The 
alteration  is,  in  fact,  nothin;jj  more  than  borrowing  10,  for 
the  sake  of  convenience,  and  then  rejecting  it  in  the  result. 

Instead  of  10,  we  may  borrow,  as  occasion  requires,  100, 
1000,  &c. 

Thus  a— fc  =  100— /;  +  a- 100  =  1 000— i  +  «— 1000,  &c. 

54.  The  DIFFERENCE  between  a  given  number  and  10,  or 
100,  or  1000,  4^c.  is  called  the  arithmetical  complement 
of  that  number. 


COMPLEMENT.  J9 

The  arithmetical  complement  of  a  number  consisting:  of 
Gjie  integral  figure,  either  with  or  without  decimals,  is  found, 
by  subtracting  the  number  from  10.  If  there  are  izvo  inte- 
gral figures,  they  are  subtracted  from  100  ;  if  three,  from 
1000,  &c. 

Thus  the  arithmetical  compl't  of  3.46  is  10 — 3.46=6.54 

of  34.6  is  100—34.6=65.4 
of  346.  is  1000— 346. =654.   &c. 

According  to  the  rule  for  subtraction  in  arithmetic,  any 
number  is  subtracted  from  10,  100,  iOOO.  &;c.  by  beginning 
on  the  right  hand,  and  taking  each  figure  from  10,  after  ///- 
rreasiiicr  all  except  the  first,  by  carrying  I. 

Thus,  if  from  10.00000 

We  subtract  7.63125 


The  difference,  orarith"!  comp't  is  2.3G875,  which  is  ob- 
tained, by  taking  5  from  10,  3  from  10,  2  from  10,  4  from  10, 
7  from  10,  and  8  from  10.  But,  instead  of  taking  each  fig- 
ure, increased  by  1,  from  10;  we  may  take  it  rvithoiU  being 
increased,  from  9. 

Thus  2  from  9  is  the  same  as  3  from  10, 

3  from  9,     the  same  as  4  from  10,  kc.     Hence, 

55,  To  obtain  the  arithmetical  complemext  of  a  num- 
ber^ subtract  the  right  hand  significant  Jigurc  from  ]0,  arid  each 
of  the  other  figures  from  d.  If,  however,  there  are  ciphers 
on  the  right  hand  of  all  the  significant  figures,  they  are  to  be 
set  down  without  alteration. 

In  taking  the  arithmetical  complement  of  a  logarithm,  if 
the  index  is  negative^  it  must  be  added  to  9  ;  for  adding  a 
negative  quantity  is  the  same  as  subtracting  a  positive  one. 
(Alg.  81.)  The  difference  between — 3  and +  9.  i.-?  not  6, 
but  12. 

The  arithmetical  complement 

of  6.24897  is  3.75103  of2. 70649  is  11.29351 

of  2.98643       7.01357  of3.G4200        6.35800 

of  0.62430      9.37570  of  9.35001        0.64999 


30  COMPOUND  PROPORTION. 

56.  The  principal  use  of  the  nrilhmeticr.l  complement,  is 
in  working  proportions  by  logiuithms;  where  some  of  the 
terms  are  to  he  added,  and  one  or  more  to  be  subtracted.  In 
the  Rule  of  'Pliree  or  simple  proportion,  two  terms  are  to  be 
added,  and  from  the  sum,  the  first  term  is  to  be  subtracted. 
But  if,  instead  of  the  logarithm  of  the  tirst  term,  we  substi- 
tute its  arithmetical  complement,  this  may  be  added  to  the 
sum  of  the  other  two,  or  more  simply,  all  three  may  be  ad- 
ded together,  by  one  operation.  After  the  index  is  diminish- 
ed by  10,  the  result  will  be  the  same  as  by  the  common  meth- 
od. For  subtracting  a  number  is  the  same,  as  adilinji;  its 
arithmetical  complement,  and  then  rejecting  10,  100,  or 
1000,  from  the  sum.  (Art.  53.) 

It'willgenerally  be  expedient,  to  place  the  terms  in  the  same 
order,  in  which  they  are  arranged  in  the  statement  of  the  pro- 
portion. 

I.  As     C'i73  «.  c.  6.20252  2.   As     253  a,  c    7.59688 

Is  to  769.4  2.88615  Isto672.5  2.82769 

So  is  37.61  1.67530  So  is  497  2.69636 


To     4.613  0.G6397  To  1321.1        3.12093 


As     46.34  a.  c.  8.33404  4     As     9.85  a.  c.  9.00656 

Is  to  892.1  2  95041  Is  to  643  2.80821 

So  is  7.633  0.88298  So  is  76.3  1 .88252 


To      147  2. 1 674:3  To  4981  3.69729 


Compound  Piioi'ORTioN 

57,  In  compound,  as  in  single  proportion,  th<^  term  !<'- 
quired  may  be  found  by  logarithms,  if  we  substitute  addition 
for  multiplication,  and  subtraction  for  division. 

Ex.  1.  If  the  interest  of  $365,  for  3  years  and  9  months, 
be  g82.13  ;  what  will  be  the  interest  of\$8940,  for  2  years 
and  6  months  ? 

In  common  arithmetic,  the  statement  of  the  question  is 
made  in  this  manner, 

*  See  VVebl.ier's  Arithmetic. 


C03irOUND  PROPORTION.  31 

.365  dollars   }  .    o-,  ^  ^    a   m       -  ■     S  8940  dollars  ; 
^^.  }    :  82.13   dollars.,     s      r^  t,  t    ' 

3. /o  years    )  (      2.5  years  ) 

And  the  method  of  calculation  is,  to  divide  the  product  ol" 
the  third,  fourth,  and  tilth  terms,  by  the  product  of  the  two 
first.*  This,  if  lo^^arithms  are  used,  will  be  to  subtract  the 
^um  of  the  logarithms  of  the  two  first  terms,  from  the  sum  of 
the  logarithms  of  the  other  three. 

.p        ..    ,   .  (  365  log.  2.56229 

Iwo  hrst  terms    ^  ^  --  n  --?^no 

(  3,  to  O.o7403 

Sum  of  the  logarithms  3.13632  j 


Third  term            82.13  1.91450 

r       ,,        1  ^A,   ,           ^    8940  3.95134 

lourthand  htth  terms  <       ^  r  /%  oo-o  , 

2.5  0.39/94 


Sum  of  the  logs,  oflhr  3d,  4th,  and  5th  G. 26378 
Do.    '  1st  and  2d     3.13632 


Term  required  1341         3.12746 


58.  The  calculation  will  be  more  e-i.nple,  if,  instead  of 
.subtracting  the  logarithms  of  the  two  first  terms,  we  add  their 
arithmetica I,  complements.  But  it  must  be  observed,  that  eac/i 
arithmetical  complement  increases  the  index  of  (he  logarithm 
by  10.  If  the  arithmetical  complement  be  introduced  into 
tioo  of  the  terms,  the  index  of  the  sum  of  the  logarithms 
will  be  20  too  great ;  if  it  be  in  three  terms,  the  index  will  be 
30  too  great,  &ic. 


^  365 
^3.75 


,,,       ^     ,    .  J  ..^..     a.  c.  /.43771 

Iwo  hrst  terms     <  .  -,_  o  .  ^^n-; 

^o    a.  c.  9.42o97 

Third  term  82  13  1.91450 

V       tu        1  ^-Ai  ^  S^^^^  3.95134 

b  ourth  and  hfth  terms  <     ^.5  0.39794 


Term  required  1341  23.12746 


The  result  is  the  sjme  as  before,  except  that  tiie  index  of 
the  logarithm  is  20  too  great. 


5i;  COMPOUND  INTEREST. 

Ex.  2,   lilhe  wages  of  53  men  for  42.  days  be  2200  dol- 
lars ;  what  will  be  the  wages  of  87  men  for  34  days  ? 

53  men  >    .  com-'  J  ^"^  ^^^l 
42  days  5    '  --^^--  ^  34  days  J 

rp        r    ,  .  (^  53  a,  c.    S.27572 

1  wo  First  terms    •;   ,^  „  c,-,^-,- 

(42  a,   c,    8.37G70 

Third  term  2200  3.34242 


terms      < 


Fourth  and  fifth  terms      ^  i\^14r 


Term  required  2923.5  3.46589 


59.  In  the  same  manner,  if  the  product  of  any  number  oi 
quantities,  is  to  be  divided,  by  theproduct  of  several  others; 
we  may  add  together  the  logarithms  of  the  quantities  to  be 
divided,  and  the  arithmetical  complements  of  the  logarithms 
of  the  divisors. 

Ex.  If  29.G7X  346.2  be  divided  by  G9.24  X7.862  X497  ; 
what  will  be  the  quotient  ? 

Numbers  to  be  divided 

Divisors 


1.47232 
2.53933 


(  29.67 

(  346.2 

C  69.24  a.  c,  8.i5964 

^  7.862   a.  c.  9.10447 

(     497  a,  c.  7.30364 


Quotient  0.03797  8.5794 


In  this  way,  the  calculations  in  Conjoined  Proportion  may 
be  expeditiously  performed. 


CoiiPouxD   Interlst. 

GO.  In  calculating  compound  interest,  the  amount  for  the 
first  year,  is  made  the  principal  for  the  second  year  ;  the 
amount  for  the  second  year,  the  principal  for  the  third  year, 
<i:c.  Now  the  amount  at  the  end  of  each  year,  must  be  pro- 
portioned to  the  principal  at  the  begiiining  of  the  year.      If 


COMPOUND   INTEREST.  33 

the  principal  for  the  first  year  be  1  dollar,  and  if  the  amount 
of  1  dollar  for  1  year  =a  ;  then,  (Alg.  377.) 

fa    :  a-  =the  amount  for  the  2d  year,  or  the  princi- 
I  pal  for  the  3d  ; 

,  .     ..    [a^  :  a^={he  amount  tor  the  3d  year,  or  the  prin- 
*  "'  '   i'  cipai  for  the  4lh  ; 

a3  :  «4_thj^.  amount  for  the  4th  year,  or  the  prin- 
cipal for  the  oth. 


I' 


That  is,  the  amount  of  1  dollar  for  any  number  of  years  is 
obtained,  by  finding  the  amount  for  1  year,  and  involving  this 
to  a  power  whose  index  is  equal  to  the  number  of  years. 
And  the  amount  of  any  other  principal,  for  the  given  time, 
is  found,  by  multiplying  the  amount  of  I  dollar,  into  the 
number  of  dollars,  or  the  fractional  part  of  a  dollar. 

If  logarithms  are  used,  the  multiplication  required  here 
may  be  performed  by  addition^  and  the  involution^  by  multi- 
plication.  (Art.  45.)     Hence, 

61.  To  calculate  Compound  Interest,  Find  the  amount  of 
1  dollar  for  1  year;  multiply  its  logarithm  by  the  number  of 
years  ;  and  to  the  product,  add  the  logarithm  of  the  principal. 
The  sum  will  be  the  logarithm  of  the  amount  for  the  given 
time.  From  the  amount  subtract  the  principal,  and  the  re- 
mainder will  be  the  interest. 

If  the  interest  becomes  due  half  yearly  or  quarterly  ;  find 
the  amount  of  one  dollar,  for  the  half  year  or  quarter,  and 
multiply  the  logarithm,  by  the  number  of  half  years  or  quar- 
ters in  the  given  time. 

If  P=thc  principial, 

a  =  the  amount  of  1  dollar  for  1  year, 
?i=any  number  of  years,  and 

A  =  the  amount  of  the  given  principal  for  n  years;  thew 
A=a"  XP. 

Taking  the  logarithms  of  both  sides  of  the  equation,  and 
reducing  it,  so  as  to  give  the  value  of  each  of  the  four  quan- 
tities, in  terms  of  the  others,  we  have 


34  COMPOUND  INTEREST. 

1.  Log.   A=»x  log.  a-r  log.  P. 

2.  Log.  P  =  logA  — ?iX  log.  «. 

log.  A  — log.  P. 

3.  Log.  a  =  — 

log.  A  — log.  P. 
log.  a. 

Any  three  of  these  quantities  being  given,  the  fourth  may 
he  found. 

Ex.  L    What  is  the  amount  of  20  dollars,  at  6  per  cent 
compound  interest,  for  100  years  ? 

Amount  ot  1  dollar  for  1  year      1.06      log.  0.0253059 
Multiplying  by  100 

2.53059 
Given  principal  20  1.30103 


Amount  required  ;^6786  3.83162 


2.  What  is  the  amount  of  I  cent,  at  6  per  cent  compound 
interest,  in  500  years? 

Amount  of  1  dollar  for  1  year  1.06         log.  0.0253059 
Multiplyirig  by  500 


12.65295 
Given  principal  0.01  -2.00000 


Amount  ^44,973,000,000  10.65295 


More  exact  answers  may  be  obtained,  by  using  logarithms 
of  a  greater  number  of  decimal  places. 

3.  What  is  the  amount  of  1000  dollars,  at  6  per  cent  com- 
pound interest,  for  10  years  ?  ^Ans.  1790.80. 

4.  What  principal,  at  4  per  cent,  interest,  will  amount  to 
1643  dollars  in  21  years  ^  Ans.  721. 


INCREASE  OF  POPULATION.  .3-, 

o.  What  principal,  at  6  per  cent,  will  amount  to  202  dol- 
lars in  4  years  ?  Ans.  160. 

6.  At  what  rate  of  interest,  will  400  dollars  amount  to 
569 J,  in  9  years  ?  Ans.  4  per  cent. 

7.  In  how  many  years  will  500  dollars  amount  to  900 
at  5  per  cent,  compound  interest  ?  Ans.  12  years. 

8.  In  what  time  will  10,000  dollars  amount  to  16,288,  at 
5  per  cent,  compound  interest  ?  Ans.  10  years. 

9.  At  what  rate  of  interest,  will  11,106  dollars  amount  to 
20,000  in  15  years?  Ans.  4  per  cent. 

10.  What  principal,  at  6  per  cent,  compound  interest, 
will  amount  to  3188  dollars  in  8  years  ?  Ans.  ^2000. 

11.  AVhat  will  be  the  amount  of  1200  dollars,  at  6  per 
cent,  compound  interest,  in  10  years,  if  the  interest  is  con- 
verted inio  principal  every  half-year? 

Ans.  2167.3  dollars. 

12.  In  what  time  will  a  sum  of  money  double,  at  G  pei 
cent,  compound  interest?  Ans.  11.9  years. 

13.  What  is  the  amount  of  5000  dollars,  at  G  per  cent, 
compound  interest,  for  28  j  years  ?  Ans.  25,942  dollars. 


Increase  of  Population. 

61.  h.  The  natural  increase  of  population  in  a  country, 
may  be  calculated  in  the  same  manner  as  compound  interest; 
on  the  supposition,  that  the  yearly  rate  of  increase  is  regu- 
larly proportioned  to  the  actual  number  of  inhabitants. 
From  the  population  at  the  beginning  of  the  year,  the  rateoi 
increase  being  given,  may  be  computed  the  whole  increase 
during  the. year.  This  added  to  the  number  at  the  begin- 
ning, will  give  the  amount,  on  which  the  increase  of  the 
second  year  is  to  be  calculated,  in  the  same  manner  as  the 
first  year's  interest  on  a  sum  of  money,  added  to  the  sum  it- 


.3ti  LNCKEASE   OF  POPULATION. 

self,  gives  the  amount  on  which  the  interest  for  the  second 
year  is  to  be  calculated. 

If  P  =  the  population  at  the  beginning  of  the  year, 

a  =  l  -4-the  fraction  which  expresses  the  rate  of  increase, 
7i=any  number  of  years  ;  and 

A=the  amount  of  the  population  at  the  end  of  n  years; 
then,  as  in  the  preceding  article, 

A=«"XP,  and 


1.  Log.  A=nXlog.  «-flog.  P. 

2.  Log.  P=log.  A  — ?iXlog.  a, 
^  loj:.  A-log.  P. 


4. 


n 
\o^.  A— log.  P. 

loiij.  a 


Ex.  1.  'J'he  population  of  the  United  States  in  1820  was 
9,625,000.  Supposing  the  yearly  rate  of  increase  to  be  ^V^h 
part  of  the  whole,  what  will  be  the  population  in  1830  ? 

Here  P=9,625,0C0.     7i=lO.     a  =  l+-U^||. 

And  log.  A  =  l0xlog.  ff  +  log.  (9,625,000,) 
Therefore,  A  =  12,060,000,  the  population  in  1830. 

2.  If  the  number  of  inhabitants  in  a  country  be  five  mil- 
lions, at  the  beginning  of  a  century  ;  and  if  the  yearly  rate 
of  increase  be  ^V  '•>  what  will  be  the  number,  at  the  end  of 
the  century?  Ans.  132,730,000. 

3.  If  the  population  of  a  country,  at  the  end  of  a  centur}^, 
is  found  to  be  45,800,000;  and  if  the  yearly  rate  of  increase 
has  been  j^o  ;  what  was  the  population,  at  the  commence- 
ment of  the  century  ?  Ans.  20  millions. 

4.  The  population  of  the  United  States  in  1810  was 
7,240,000;  in  1820,  9,625,000.  What  was  the  annual  rate 
of  increase  between  these  two  periods,  supposing  the  in- 
crease each  year  to  be  proportioned  to  the  population  at  the 
beginning  of  the  year? 


IN'CREASE   OF  POPULATION.  37 

loir.  9,625,000  — locr.  7.240,000 
Here  log.  «=— ^- rr: 

Therefore,  «=1.029;  and  yff^,  or  2.9  percent,  is  the 
rale  of  increase. 

5.  In  how  many  years,  will  the  population  of  a  i.ountry 
advance  from  two  millions  to  five  millions;  supposing  the 
yearly  rate  of  inciease  to  be  3I0  ?  Ans.  47|  years. 

6.  If  the  population  of  a  country,  at  a  given  time,  be  seven 
n^.illions ;  and  if  the  yearly  rate  of  increase  be  ^\{h  ;  what 
will  be  the  population  at  the  end  of  35  years  ? 

7.  The  population  of  the  United  States  in  1800  wa> 
5,306,000.  VVhat  was  it  in  1780,  supposing  the  yearly  rate 
of  increase  to  be  ^V ' 

8.  In  what  time,  will  the  populalioii  of  a  country  advance 
from  four  millions  to  seven  millions,  if  the  latio  of  increase 
be  -^- ? 

9.  What  must  be  the  rate  of  increase,  that  the  population 
of  a  place  may  change  from  nine  thousand  to  fifteen  thou- 
sand, in  12  years  ? 

If  the  population  of  a  country  is  not  aiPected  by  immigra- 
tion or  emigration,  the  rate  of  increase  will  be  equal  to  the 
ditference  between  the  ratio  of  the  ^irM^,  and  the  ratio  of  the 
deaths,  when  compared  with  the  whole  population. 

Ex.  10-  If  the  population  of  a  country,  at  any  given  time, 
he  ten  millions  ;  and  the  ratio  of  the  armual  number  of  births 
to  the  whole  population  be  ^\,  and  the  ratio  of  deaths  jj, 
what  will  hi  the  number  of  inhabitants^,  at  the  end  of  60 
years  ? 

Here  in?  yearly  rate  of  increase  =  VT  —  t    =T^n. 

And  the  population,  at  the  end  of  60  year?=  31,750,000. 

The  rate  of  increase  or  decrease  from  immig ration  or  em- 
igration, will  be  equal  to  the  ditrorence  between  the  ratio  of 
immigration  and  the  ratio  of  emigration  ;  and  if  this  differ- 


r?«>Or' 


38  INCREASE   OF  POPULATION. 

ence  be  added  to,  or  subtracted  from,  the  difference  between 
the  ratio  of  the  births  and  that  of  the  deaths,  the  whole  rata 
of  increase  will  be  obtained. 


Ex.  11.  If  in  a  country,  the  ratio  of  births  be 


3  01 

the  ratio  of  deaths  ^^j 


a  oi 
_J_. 

G  01 


the  ratio  of  immigration     j\ 
the  ratio  of  emigration 

and  if  the  population  this  year  be  10  millions,  what  will   it 

be  20  years  hence  ? 

The  rate  of  the  natural  increase     =-^^__i_  =  _i_  j 
That  of  increase  from  immigration  =  5V~  6'o  =  3io  ; 


The  sum  of  the  two  is  -- 


6  0  0? 


And  the  population  at  the  end  of  20  years,  is  12,611,000. 

1  2.  If  the  ratio  of  the  births  be         -^'oi 

of  the  deaths,  3V, 

of  immigration,         j\, 

of  emigration,  ~, 

in  what  time  will  three  millions  increase  to  four  and  a  half 
millions  ? 

If  the  period  in  which  the  population  will  double  be  given; 
the  numbers  for  several  successive  periods,  will  evidently 
be  in  a  geometrical  progression,  of  which  the  ratio  is  2;  and 
as  the  number  of  periods  will  be  one  less  than  the  number  of 
terms: 

If  P=the  first  term, 

A=thc  last  term, 

n=the  number  of  periods  ; 
Then  will  A  =  PX2",  (Alg.  439.) 
Or  log.  A  =  log.  P.  =  log.  P-{-nXlog.  2. 

Ex.  1.  If  the  descendants  of  a  single  pair  double  once  in 
25  years,  what  will  be  their  number,  at  the  end  of  one  thou- 
sand years  ? 

The  number  of  periods  here  is  40. 
And  A  =  2X2^  "  =  2.199.200.000,000. 


EXPONENTIAL  EHUATIONS.  ^9 

■2.  If  the  descendants  of  Noah,  beginning  with  his  three 
sons  and  their  wives,  doubled  once  in  20  years  for  300  years  ; 
what  was  their  number,  at  the  end  of  this  time  ? 

Ans.   196,608. 

3.  The  population  of  the  United  States  in  1820  being 
9,625,000  ;  what  must  it  be  in  the  year  2020,  supposing  it  to 
double  once  in  25  years  ?  Ans.  2,464,000,000. 

4.  Supposing  the  descendants  of  the  first  human  pair  to 
double  once  in  50  years,  for  1650  years,  to  the  time  of  the 
deluge,  what  was  the  population  of  the  world,  at  that  time  ? 


Exponential  Equations. 

62.  An  Exponential  equation  is  one  in  which  the  letter 
expressing  the  unknown  quantity  is  an  exponent. 

Thus  cf=^b,  and  3c'=bc,  are  exponential  equations.  These 
are  most  easily  solved  by  logarithms.  As  the  two  members 
of  an  equation  are  equal,  their  logarithms  must  also  be  equal. 
If  the  logarithm  of  each  side  be  taken,  the  equation  may 
then  be  reduced,  by  the  rules  given  in  algebra. 

Ex,  What  is  the  value  of  x  in  the  equation  3^=243? 

Taking  the  logarithms  of  both   sides    log.    (3*)=log.  243 
But  the  logarithm  of  a  power  is  equal  to  the  logarithm  of 
the  root,  multiplied  into  the  index  of  the  power.  (Art.  45.) 

Therefore  (log.3)Xi=log.  243  ;  and  dividing  by  log.  3, 
log.  243     2.38561 
^=l^i7r=o:^i2  =  5-     So  that  3^  =243. 

63.  The  preceding  is  an  exponential  equation  of  the  sim- 
plest  form.  Other  cases,  after  the  logarithm  of  each  side  is 
taken,  may  be  solved  by  Trial  and  Err  our  ^  in  the  same  man- 
ner as  affected  equations.  (Alg.  503.)  For  this  purpose, 
make  two  suppositions  of  the  value  of  the  unknown  quantity, 
and  find  their  errours  :  then  say, 


40  EXPONENTIAL 

As  the  dilFerence  of  the  crroiirs,  to  the  dif- 
ference of  the  assumed  numbers  ; 

So  is  the  least  errour,  to  the  correction  required 
in  the  corresponding  assumed  number. 

Ex.  1.  Find  the  value  ofx  in  the  equation  x"  =256 
Taking  the  logarithms  of  both  sides     (log.  a[;)Xa;  =  log.  256 
Let  X  be  supposed  equal  to  3.5,  or  3.6. 

By  the  first  supposition.  liy  the  second  supposition . 

.x  =  3.5,  and  log.  a;  =  0.54407       a=3.6,  and  log.  a;  =  0.55630 
Multiplying  by  3.5  Multiplying  by  3.6 

(log.  ?;)Xx=l. 90424  (log.  a^)  X.r  ==2.00268 

log.        25G  =  2.40ry24  log.        256=2.40824 

Errour         —0.50400  Errour  —0.40556 

Difference  of  the  errours  0.09844 

Then  0.09844  :  0.1  :  10.40556  :  0.41 19,  the  correction. 
This  added  to  3.6,  the  second  assumed  number,  makes  the 
value  of  x=4.0119. 

To  ct)rrect  this  farther,  suppose  a  =  4.01 !,  or  4.012. 

By  the  first  supposition.  By  the  second  supposition. 

.i=4.0n,andlog.  a:  =  0.60325  a;=4.012,and  log.  a;  =  0.60336 

Multiplying  by         4.011  Multiplying  by  4.012 

(log.  x)Xx=2.41963  (log.  a:)  X a;  =2.42068 

log.      256=2.40824  log.         256=2.40824 


Errour       +0.01139  Errour     +0.01244 

Difference  of  the  errours  0.00105 

Then  0.00105  :  0.001 : !  0.01 139  :  0.011  very  nearly. 
Subtracting  this  correction  from  the  first  assumed  number 
4.011,  we  have  the  value  of  x=4,  which  satisfies  the  condi- 
tions of  the  proposed  equation  ;  for  4*  =256, 

2.  Reduce  the  equation  4a:*  =100a:^.  Ans.  x  =  5. 

3.  Reduce  the  t-quaiion  a  -  =9,t. 


EQUATIONS.  41 

04.  The  exponent  of  a  power  may  be  itself  a  power,  as  in 
the  equation 

a^"  =6. 
where  x  is  the  exponent  of  the  power  nf  ,  which  is  the  expo- 
nent of  the  power  a"'""  . 

Ex.  4.  Find  the  ralue  of  x,  in  the   equation  9^   =1000. 

3^  X(log.  9)  =  log.  1000.     Therefore  3'  =^^1^:9  =3.14 

log.  9 
Then  as  3    =3.14.     x  (log.  3)  =  log.  (3.  14.) 

Therefore  x=^'^^^d^Ll^-^:\^\%\^  =  \.OA. 
log.  3. 

In  cases  like  this,  wliere  the  factors,  divisors,  &c.  are  loga- 
rithms, the  calculation  may  be  facihtated,  by  taking  the  /oo-- 
arithms  of  the  logarithms.  Thus  the  value  of  the  fraction 
•HttIt! 's  most  easily  found,  by  subtracting  the  logarithm  of 
the  logarithm  which  constitutes  the  denominator,  from  the 
logarithm  of  that  which  forms  the  numerator. 

Find  the  value  of  x,  in  the  equation  ^^"^  +</=m. 

c 

Ans.  x=]£Li£!?ZI^!£ili- 
Jog. «, 


SECTION  IV. 

DIFFERENT  SYSTEMS  OF  LOGARITHMS,  AND 
COMPUTATION  OF  THE  TABLES. 

„  ^  Tj^OR  the  common  purposes  of  numerical  computation, 
^*  Briggs' system  of  logarithms  has  a  decided  advantage 
overevery  other.  Butthe  theory  of  logarithms  is  an  important 
instrument  of  investigation,  in  the  higher  departments  of 
mathematical  science.  In  its  numerous  applications,  there 
is  frequent  occasion  to  compare  the  common  system  with  oth- 
ers ;  especially  with  that  which  was  adopted,  by  the  cele- 
brated inventor  of  logarithms,  Lord  Napier.  In  conducting 
these  investigations,  it  is  often  expedient  to  express  the  loga- 
rithm of  a  number,  in  the  form  of  a  series. 

If  a  ^=N,  then  x  is  the  logarithm  ofN.  (Art.  2.) 
To  find  the  value  of  a?,  in  a  series,  let  the  quantities  a  and 
N  be  put  into  the  form   of  a  binomial,  by  making  «  =  l-h6, 
and  N  =  l+?i.     Then  {\-\-bY  =(14-n),   and   extracting  the 
root^  of  both  sides,  we  have 

By  the  binomial  theorem 

-f  &c. 

y  y    y  V  2/      2/^y       ^^y  \2.3/ 

-I-&C. 

As  these  expressions  will  be  the  same,  whatever  be  the 
value  of  !/,  let  y  be  taken  indefinitely  great  ;  then  ^  and  ^ 
being  indefinitely  small,  in  comparison  with  the  numbers  -  1, 
—  2,  &c.  with  which  they  are  connected,  may  be  cancelled 
(Alg.  456.)  leaving 

■+j'-i(n+K?)-i(n^--+.'"-i(?) 


DIFFERENT  SYSTEMS  OF  LOGARITHMS.  43 

Rejecting  1  from  each  side  of  the  equation,  nrmltiplyin?  by 
y,  and  dividing  by  the  compound  factor  into  which  x  is  mul- 
tiphed,  we  have 

Or,  as  71  =  N—  1,  and  6=a  —  1, 

^'  (a-l)-i(a-l)2  +  i(a-l)^-i(a-l)^+&c. 

Which  is  a  general  expression,  for  the  logarithm  of  any 
number  N,  in  any  system  in  which  tlie  base  is  n.  The  nu- 
merator is  expressed  in  terms  of  N  only  ;  and  the  denomi- 
nator in  terms  of  a  only  :  So  that,  whatever  be  the  number, 
the  denominator  will  remain  the  same,  unless  the  base  is 
changed.     The  reciprocal  of  this  constant  denominator,  viz. 

1 


(a-l)-H«-l)=-|-H«-0^-i(«-  ly+^c. 
is  called  the  Modulus  of  the  system  of  which  a  is  the  base.  II 
this  be  denoted  by  M,  then 

Log.  N=Mx((N-l)-i(N-l)-^-fKN-l)='-i(N-l)* 

66.  The  foundation  of  Napier's  system  of  Logarithms  is  laid, 
by  making  the  modulus  equal  io  unity.  From  this  condition 
the  6a5e  is  determined.  Taking  the  equation  marked  A.  and 
making  the  denominator  equal  to  1,  we  have 

By  reverting  this  equation* 

2        2.3         2.3.4       2.3.4.5 
Or,  as  by  the  notation,  n-|-l=N=a% 

2  ^  2.3  ^2.3.4^2.3.4.5 
If  then  X  be  taken  equal  to  1,  we  have 

a=  1 -f  I +.i_ -|-_L  4-_L  __  &c. 
2.3      2.3.4     2.3.4.5 

Adding  the  first  fifteen  terms,  we  have 

2.7182818284 
Which  is  the  base  of  Napier's  system,  correct  to  ten  pla- 
ces of  decimals. 

*  See  note  D. 


44         DIFFERENT  SYSTEiMS  OF  LOGARITHMS. 

Napier's  logarithms  are  also  called  hijperbolic  logarithms, 
from  certain  relations  which  they  have  to  the  spoces  between 
the  asymptotes  and  the  curve  of  an  hyperbola  ;  although 
these  relations  are  not,  in  fact,  peculiar  to  Napier's  system. 

67.  The  logarithms  o(  different  systems  are  compared  with 
each  other,  by  means  of  the  modulus.     As  in  the  series 

(N-l)-i(N-1)^-f-KN-l)^-i(N-l)^4-&c. 

which  expresses  the  logarithm  of  N,  the  denominator  only  is 
affected  by  a  change  of  the  base  «;  and  as  the  value  of  frac- 
tions, whose  numerators  are  given,  are  reciprocally  as  their 
denominators.    (Alg.  360.  cor.  2.) 

The  logarithm  of  a  given  number,  in  one  system, 

Is  to  the  logarithm  of  the  same  number  in  another  system; 

As  the  modulus  of  one  system^ 

To  the  modulus  of  the  other. 

So  that,  if  the  modulus  of  each  of  the  systems  be  given, 
and  the  logarithm  of  any  number  be  calculated  in  one  of 
the  -ystems;  the  logarithm  of  the  snme  number  in  the  other 
system  may  be  calculated  by  a  simple  proportion.  Thus  if 
M  be  the  modulus  in  Brigg's  system,  and  M'  the  modulus  in 
Napier's  ;  /  the  logarithm  of  a  number  in  the  former,  and  /' 
the  logarithm  of  the  same  number  in  the  latter ;  then, 

jM  :  M'::/:/', 

Or,  as  M'  =  l 

M:i::/:/' 

Therefore,  /=/'xM,  that  is,  the  common  logarithm  of  a 
number,  is  equal  to  Napier's  logarithm  of  the  same,  multi- 
plied 'n^o  the  modulus  of  the  common  system. 

To  find  this  modulus,  let  a  be  the  base  of  Brigg's  system, 
and  e  ^he  base  of  Napier's  ;  and  let  La  denote  the  common 
logarithm  ofa,  and  /'.a  denote  Napier's  logarithm  of  a. 

Then  M  :  1 :  :/.a  :  I'a.  Therefore  U—ir- 

But  in  the  common  system,  a=10,  and  /.a  =  l. 

J 
So  that,   M.=YTc)^  ^^^^  '^'  *^^  modulus  of  Brigg's  system, 

is  equal  to  1  divided  by  Napier's  logarithm  of  10. 


COMPUTATION  OF   LOGARITHMS.  45 

Again  M  :  i::Le  :  I'.e 

But  as  e  denotes  Napier's  base,  /'.e  =  l. 
So  that  M  =  /.e,  that  \<,  the  modulus  of  the  common  sys- 
tem, is  equal  to  the  common  logarithm  of  Napier's  base. 

Therefore,  either  of  the  expressions,  /.e,  or  ^7- ,  may  be 

used,  to  convert  tlie  logarithms  of  one  of  the   systems   into 
those  of  the  other. 

The  ratio  of  the  logarithms  of  two  numbers  to  each  other^  is 
the  same  in  one  system  as  in  another.  If  N  and  n  be  the  two 
numbers  ; 

Then,  /.N  :  /'.N::M  :  M' 

InWV.n'.'M  :  xM' 
Therefore,  /.N  :  /.n::/'.N  :  l\n 


Computation  of  Logarithms. 

68.  The  logarithms  of  most  numbers  can  be  calculated  by 
approximation  only,  by  finding  the  sum  of  a  sufficient  number 
of  terms,  in  the  series  which  expresses  the  value  of  the  loga- 
rithms.    According  to  art.  Qb. 

Log.  N  =  MX(  (N-l)-i(N-l)^-f-i(N-l)%  &c.) 
Or,  putting  as  before,  7i  =  N—  1, 
Log.  (]-f-n)  =  M(/i-iH^+in3-in*+in^— &c.) 
But  this  series  will  not  converge,  when  n  is  a  whole  num- 
ber, greater  ihan  unity.     To  convert  it  into  another  which 
will  converge,  let  (1  —n)  be   expanded  in  the   same  manner 
as  (1-f-^),  (Art.  65.)     The  formula  will  be  the  same,  except 
that  the  odd  powers  of  n  will  be  negative  instead  of  positive. 
We  shall  then  have. 

Log.  (l4-w)  =  M(n-^w2-fin.3-in*-l-in^-&c.) 
Log.  (l-n)  =  M(-n-iM2^in='-in»-fin5~&c.) 

Subtracting  the  one  from  the  other  the  even  powers  of  n 
disappear,  and  we  have 

M(2yi-f|n34-fnS+fn^-{-&£C.) 

or 
2M  (n-f-in'' 4-1^5 -fi/i^+Stc.) 


46  COMPUTATION    OF  LOGARITHMS. 

But  this,  which  is  the  difference  of  the  logarithms  of  (l  +?^) 
and  (1— n)  is  the  logarithm  of  the  quotient  of  the  one  divided 
by  the  other.   (Art.  36.) 

1-l-n 
That  is,  Log.  j^:;;^=2M(n4-^n3-f  iw^-f  in-^H-^c.) 

1 
Now  put  n= 

^  z  —  1 

1  z 

1  + 


Then 


'      \-~n     ^_J ■g-2 

z-\      z-\ 


z               1+n          ,     1 
Therefore,  substitutina;    ^  for   :; ,  and :     for    w. 

'  ^  z—2  1  —n         'z—  1 

we  have 

Log.  ,4i=2M((^4:yy4-5^34-^(I^ 
Or,  (Art.  36,) 

Log..-log.(.-2)  =  2M((^y+5^,+^^ 
Therefore, 

Log.  z=iog.  {z-2)^m{j~^~:^^ 

This  series  may  be  applied   to  the   computation  of  any 
number  greater  than  2. 

To  find  the  logarithm  of  2,  let.  ^  =  4, 
Then  (r  — 1)  =  3,  and  the  preceding  series,  after  transpos- 
ing log.  {z  —  2)  becomes 

Log.  4-log.2=2M(3+5;^+^;^+7;^<Sic.J 

But  as  4  is  the  square  of  2  log.  4  =  2log.  2.   (Alg.  44-)    So 
that  log.  4  — log.  2  =  log.  2.     We  have  then 


NAPIER'S  OR  HYPERBOLIC  LOGARITHMS.         47 
/I         1  1  1      ,      1  ^     \ 

Log.  2=2M  [s+^:^+jj^+y:^'^jj^-^^<^') 

When  the  logarithms  of  the  prime  numbers  are  computed, 
the  logarithms  of  all  other  numbers  may  be  found,  by  sim- 
ply adding  the  logarithms  of  the  factors  of  which  the  num- 
bers are  composed.   (Art.  36.) 

60.  In  Napier's  system,  where  M  =  l,the  logarithms  may 
be  computed,  as  in  the  following  table. 

Napier's  or  Hyperbolic    Logarithms. 

Log.  2=2  (3+.^:^+^;^+^r^&c.)  =0.693147 

Log.  3  =  2  (i-|.-l-+-L_4.^  &c.)  =L0986I2 

Log.  4  =  2  log.  2  1.386294 

Log.  3=log.  3  +  2  (i+5^+5^+:^4rc.)    =1.609438 

Log.  6=log.  3-f  log.  2.  =1.791759 

Log.  7  =  log.  5  +  2(^+^4-^4--^ &ic.)  =1.955900 

Log.  8  =  log.  4+log.  2.  =2.079441 

Log.  9  =  2  log.  3.  =2.197224 

Log.  10=log.  5  +  log.  2.  =2.302585 

he.  &c.  &c. 

70.  To  compute  the  logarithms  of  thecommon  system,  it  will 
be  necessary  to  find  the  value  of  the  modulus.     This  is  equal 
to  1  divided  by  Napiers'  logarithm  of  10  (Art.  67.)  that  is, 
_1 

2.302585  =  . 43429448. 

This  number  substituted  for  M,  or  twice  the  number  viz. 
.86858896  substituted  for  2  M,  in  the  series  in  art.  68.  will 
enable  us  to  calculate  the  common  logarithm  of  any  number. 


48  COMMON  OR  BRIGG'S  LOGARITHMS. 

Common  or  Brigg's  Logarithms. 

Log.  2=.86858896(--f  ^+^-^-|-y^~&c.)  =0.301030 

Log.  3  =  86858896  (-+^--f--|^+^2,  &c.)  =0.477121 

Log.  4=2  log.  2.  =0.602060 

Log.  5=log.  10  — log.  2=1  —log.  2.  =0.698970 

Log.  6  =  log.  3.-f  log.  2.  =0.778151 

Log.  7  =  86858896(^+^+^+^&c.) 

+  log.  5  =0.855098 

Log.  8  =  3  log.  2.  =0.903090 

Log.  9=2  log.  3.  =0.954243 

Log.  10.  =1.000000 

&c.  &c. 


TRIGOXOMETRVc 


SECTION  I. 


SINES,  TANGENTS,  SECANTS,  &c. 

sides  and  angles  of  TRiAiiGLES,  J ts  first  object 
is,  to  determine  the  length  of  the  sides,  and  the  quantity  of 
the  angles.  In  addition  to  this,  from  its  principles  are  deriv- 
ed many  interesting  methods  of  investigation  in  the  higher 
branches  of  analysis,  particularly  in  physical  astronomy. 
Scarcely  any  department  of  mathematics  is  more  important, 
or  more  extensive  in  its  applications.  By  trigonometry,  the 
mariner  traces  his  path  on  the  ocean  ;  the  geographer  deter- 
mines the  latitude  and  longitude  of  places,  the  dimensions 
and  positions  of  countries,  the  altitude  of  mountains,  the 
courses  of  rivers,  he.  and  the  astronomer  calculates  the  dis- 
tances and  magnitudes  of  the  heavenly  bodies,  predicts  the 
eclipses  of  the  sun  and  moon,  and  measures  the  progress  of 
light  from  the  stars. 

72.  Trigonometry  is  either  platie  or  spherical.  The  for- 
mer treats  of  triangles  bounded  by  right  lines  ;  the  latter,  o( 
triangles  bounded  by  arcs  of  circles. 

Divisions  of  the  Circle. 

73.  In  a  triangle  there  are  two  classes  of  quantities  which 
are  the  subjects  of  inquiry,  the  sides  and  the  angles.  For 
the  purpose  of  measuring  the  latter,  a  circle'is  introduced. 

The  periphery  of  every  circle,  whether  great  or  small,  is 
supposed  to  be  divided  into  360  equal  parts  called  degrees, 
each  degree  into  60  minntes,  each  minute  into  60   seconds, 

8 


50  TRIGONOMETRY 

each  second  into  60  thirds,  &ic.  marked  with  the  characters 
°, ',  ", '",  he.  Thus  32°  24'  1 3"  22'"  is  32  degrees  24  minutes 
13  seconds  22  thirds.* 

A  degree,  then,  is  not  a  magnitude  of  a  given  length  ;  hut 
a  certain  portion  of  the  whole  circumference  of  any  circle. 
It  is  evident,  that  the  360th  part  of  a  large  circle  is  greater, 
than  the  same  part  of  a  small  one.  On  the  other  hand,  the 
number  of  degrees,  in  a  small  circle,  is  the  same  as  in  a  large 
one. 

The  fourth  part  of  a  circle  is  called  a  quadrant,  and  con- 
tains 90  degrees. 

74.  To  measure  an  angle,  a  circle  is  so  described  that  its 
centre  shall  be  the  angular  point,  and  its  periphery  shall  cut 
the  two  lines  which  include  the  angle.  The  arc  between  the 
two  lines  is  considered  a  measure  of  the  an^le,  because,  by 
Euc.  33.  6,  angles  at  the  centre  of  a  given  circle,  have  the 
same  ratio  to  each  other,  as  the  arcs  on  which  they  stand. 
Thus  the  arc  AB,  (Fig.  2.)  is  a  measure  of  the  angle  ACB. 

It  is  immaterial  w^hat  is  the  size  of  the  circle,  provided  it 
cuts  the  lines  which  include  the  angle.  Thus  the  angle 
ACD  (Fig  4.)  is  measured  by  either  of  the  arcs  AG,  ag. 
For  ACD  is  to  ACH,  as  AG  to  AH.  or  as  ag  to  ah.  (Euc. 
33.  6.) 

75.  In  the  circle  ADGH  (Fig.  2.)  let  the  two  diameters 
AG  and  DH  be  perpendicular  to  each  other.  The  angles 
ACD,  DCG,  GCH,  and  HCA  will  be  right  angles  ;  and  the 
periphery  of  the  circle  will  be  divided  into  four  equal  parts, 
each  containing  90  degrees.  As  a  right  angle  is  suhtended 
by  an  arc  of  v^0°,  the  angle  itself  is  said  to  contain  90°.  Hence, 
in  two  right  angles,  there  are  180°,  in  four  right  angles  360°; 
and  in  any  other  angle,  as  many  degrees,  as  in  the  arc  by 
which  it  is  subtended. 

76  The  sum  of  the  three  angles  of  any  triangle  being 
equal  to  two  right  angles,  (Euc.  32.  1.)  is  equal  to  180°. 
Hence,  there  can  never  be  more  than  one  obtuse  angle  in  a 
triangle.  For  the  sum  of  two  obtuse  angles  is  more  than 
180°. 

77.  The  COMPLEMENT  of  an  arc  or  an  angle^  is  the  differ* 
ence  between  the  arc  or  angle  and  90  degrees. 

The  complement  of  the  arc  AB  (Fig.  2.)  is  DB  ;  and  the 
complement  of  the  angle  ACB  is  DCB.  The  complement 
of  the  arc  BDG  is  also  DB. 

*  See  note  E 


SINES  TANGENTS,  &c,  51 

The  complement  of  10°  is  80°,         of    60°  is  30°, 
of  20°  is  70°,  of  120   is  30°, 

of  50°  is  40°,         of  170   is  80°,  &c. 

Hence  an  acute  angle  and  its  complement  are  always  equal 
to  90°.  The  angles  ACB  and  DCB  are  together  equal  to  a 
right  angle.  The  two  acute  angles  of  a  right  angled  trian- 
gle are  equal  to  90°  :  therefore  each  is  the  complement  of  the 
other. 

78.  The  SUPPLEMENT  of  an  arc  or  an  angle  is  the  difference 
betrceen  the  arc  Granule  and  180  degrees. 

The  supplement  of  the  arc  BDG  (Fig.  2.)  is  AB  ;  and  the 
supplement  of  the  angle  BCG  is  BCA. 

The  supplement  of  10°  is  170°,         of  120°  is  60°. 

of  80°  is  100°,         of  150°  is  30°,  &tc. 

Hence  an  angle  and  its  supplement  are  always  equal  to 
180°.  The  angles  BCA  and  BCG  are  together  equal  to  two 
right  angles. 

79.  Cor.  As  the  three  angles  of  a  plane  triangle  are  equal 
to  two  r.ght  angles,  that  is,  to  180°  (Euc.  32  1.)  the  sum  of 
any  two  of  them  is  the  supplement  of  the  other.  So  that  the 
third  angle  may  be  found,  by  subtracting  the  sum  of  the  other 
two  from  1 80°.  Or  the  sum  of  any  two  may  be  found,  by 
subtracting  the  third  from  180°. 

^0.  A  straight  line  drawn  from  the  centre  of  a  circle  to 
any  part  of  the  periphery,  is  called  a  radius  of  the  circle. 
In  many  calculations,  it  is  convenient  to  consider  the  radius, 
whatever  be  its  length,  as  a  unit.  (Alg.  510.)  To  this  must 
be  referred  the  numbers  expressing  the  lengths  of  other  lines. 
Thus  20  will  be  twenty  times  the  radius,  and  0.75,  three 
fourths  of  the  radius. 

Definitions  of  Sines,  Tangents,  Secants,  dire. 

81.  To  facilitate  the  calculations  in  trigonometry,  there 
are  drawn,  within  and  about  the  circle,  a  number  of  straight 
lines,  called  Sines,  Tangents,  Secants,  ^c.  With  these  the 
learner  should  make  himself  perfectly  familiar. 

82.  TTie  Sine  of  an  arc  is  a  straight  tine  drawn  from  one 
end  of  the  arc,  perpendicular  to  a  dinmeter  which  passes  through 
fhe  other  end. 


62  TRIGONOMETRY. 

Thus  BG  (Fig.  3.)  is  the  sine  of  the  arc  AG.  For  BG  is 
a  line  drawn  from  the  end  G  of  the  arc,  perpendicular  to  the 
diameter  AM  which  passes  through  the  other  end  A  of  the 
arc. 

Cor.  The  sine  is  half  the  chord  of  double  the  arc.  The 
sine  BG  is  half  PG,  which  is  the  chord  of  the  arc  PAG, 
double  the  arc  AG. 

83.  The  VERSED  sine  of  an  arc  is  that  part  of  the  diameter 
which  is  between  the  sine  and  the  arc. 

Thus  BA  is  the  versed  sine  of  the  arc  A(t. 

84.  The  TANGENT  of  an  arc,  is  a  straight  line  drawn  per- 
pendicular from  the  extremity  of  the  diameter  which  passes 
through  one  end  of  the  arc,  and  extended  till  it  meets  a  line 
drawn  from  the  centre  through  the  other  end. 

Thus  AD  (Fig.  3.)  is  the  tangent  of  the  arc  AG. 

85.  The  Secant  of  an  arc,  is  a  straight  line  drawn  from  the 
centre,  through  one  end  of  the  arc,  and  extended  to  the  tangent 
which  is  drawn  from  the  other  end. 

Thus  CD  (Fig.  3.)  is  the  secant  of  the  arc  AG. 

81-.  In  Trigonometry,  the  terms  tangent  and  secant  have  a 
more  limited  meaning,  than  in  Geometry.  In  both,  indeed, 
the  tangent  touches  the  circle,  and  the  secant  c?/^s-  it.  But  in 
Geometry,  these  lines  are  of  no  determinate  length ;  whereas, 
in  Trigonometry,  they  exteiid  from  the  diameter  to  the  point 
in  which  they  intersect  each  other. 

87.  The  lines  just  defined  are  sines,  tangents  and  secants 
o(  arcs,  BG  (Fig.  3.)  is  the  sine  of  the  arc  AG.  But  this 
arc  subtends  the  angle  GCA.  BG  is  then  the  sine  of  the  arc 
which  subtends  the  angle  GCA.  This  is  more  concisely  ex- 
pressed, by  saying  that  BG  is  the  sine  of  the  angle  GCA. 
And  universally,  the  sine,  tangent,  and  secant  of  an  arc,  are 
said  to  be  the  sine,  tangent,  and  secant  of  the  angle  which 
stands  at  the  centre  of  the  circle,  and  is  subtended  by  the  arc. 
"Whenever,  therefore,  the  sine,  tangent,  or  secant  of  an  an- 
gle is  spoken  of;  T»'e  are  to  suppose  a  circle  to  be  drawn 
whose  centre  is  the  angular  point ;  and  that  the  lines  men- 
tioned belong  to  that  arc  of  the  periphery  which  subtends 
the  angle. 

88.  The  sine,  and  tangent  of  an  acute  angle,  are  opposite 
to  the  angle.  But  the  secant  is  one  of  the  lines  which  include 
the  angle.  Thus  the  sine  BG,  and  the  tangent  AD  (Fig.  3.) 
are  opposite  to  the  angle  DCA.  But  the  secant  CD  is  one 
of  the  lines  which  include  the  angle. 


SINES,  TANGENTS  &c.  53 

89.  The  sine  complement  or  cosine  of  an  angle,  is  the  sine 
of  the  COMPLEMENT  of  that  angle.  Thus,  if  the  diameter 
HO  (Fig.  3.)  be  perpendicular  to  MA,  the  angle  HCG  is  the 
complement  of  ACG  ;  (Art.  77.)  and  LG,  or  its  equal  CB, 
is  the  sine  of  HCG.  (Art.  82.)  It  is,  therefore,  the  cosine  of 
GCA.  On  the  other  hand  GB  is  the  sine  of  GCA,  and  the 
cosine  of  GCH. 

So  also  the  cotangent  of  an  anj^le  is  the  tangent  of  the  com- 
plement of  the  angle.  Thus  HF  is  the  cotangent  of  GCA. 
And  the  cosecant  of  an  angle  is  the  secant  of  the  complement 
of  the  angle.     Thus  CF  is  the  cosecant  ofGCA. 

Hence,  as  in  a  right  angled  triangle,  one  of  the  acute  an- 
gles is  the  complement  of  the  other  ;  (Art.  77.)  the  sine,  tan- 
gent, and  secant  of  one  of  these  angles,  are  the  cosine,  co- 
tangent, and  cosecant  of  the  other. 

90.  The  sine,  tangent,  and  secant  of  the  supplement  of  an 
angle,  are  each  equal  to  the  sine,  tangent,  and  secant  of  the 
angle  itself.  It  will  be  seen,  by  applying  the  definition  (Art. 
82.)  to  the  figure,  that  the  sine  of  the  obtuse  angle  GCM  is 
BG,  which  is  also  the  sine  of  the  acute  angle  GCA.  It  should 
be  observed,  however,  that  the  sine  of  an  acute  angle  is  op- 
posite to  it  ;  while  the  sine  of  an  obtuse  angle  falls  without 
the  angle,  and  is  opposite  to  its  supplement.  Thus  BG,  the 
sine  of  the  angle  MCG,  is  not  opposite  to  MCG,  but  to  its 
supplement  ACG. 

The  tan(rc7it  of  the  obtuse  angle  MCG  is  MT,  or  its  equal 
AD,  which  is  also  the  tangent  of  ACG.  And  the  secant  of 
MCG  is  CD,  which  is  also  the  secant  of  ACG. 

91.  But  the  versed  sine  of  an  angle  is  not  the  same,  as  that 
of  its  supplement.  The  versed  sine  of  an  acute  angle  is  equal 
to  the  difference  between  the  cosine  and  radius.  But  the 
versed  sine  of  an  obtuse  angle  is  equal  to  the  sum  of  the  co- 
sine and  radius.  Thus  the  versed  sine  of  ACG  is  AB  =  AC 
-  BC.  (Art.  83.)  But  the  versed  sine  of  MCG  is  MB  =  MC 
4-BC. 

Relations  of  Siiies,  Tangents,  Secants,  t^^c,  to  each  other. 

92.  The  relations  of  the  sine,  tangent,  secant,  cosine,  &€. 
to  each  other,  are  easily  derived  from  the  proportions  of  the 
sides  of  similar  triangles.  (Euc.  4.  6.)  In  the  quadrant  ACH, 
(Fig.  3.)  these  lines  form  three  similar  triangles,  viz.  ACD, 
BCG  or  LCG.  and  HCF.     For,  in  each  of  these,  there  is  one 


64  TRIGONOMETRY. 

right  angle,  because  the  sines  and  tangents  are,  by  definition, 
perpendicular  to  AC  ;  as  the  cosine  and  cotangent  are  to 
CH.  The  hnes  CH,  BG,  and  AD  are  parallel,  because  CA 
makes  a  right  angle  with  each.  (Euc.  27.  1.)  For  the  same 
reason,  CA,  LG,  and  HF  are  parallel.  The  alternate  anj^les 
GCL,  BGC,  and  the  opposite  angle  CDA  are  equal  ;  (Euc. 
29.  1.)  as  are  also  the  angles  GCB,  LGC,  and  HFC.  The 
triangles  ACD,  BCG,  and  HCF  are  therefore  similar. 

It  should  also  be  observed,  that  the  line  BC,  between  the 
sine  and  the  centre  of  the  circle,  is  parallel  and  equal  to  the 
cosine  ;  and  that  LC,  between  the  cosine  and  centre,  is  par- 
allel and  equal  to  the  sine  ;  (Euc.  34.  1.)  so  that  one  may  be 
taken  for  the  other,  in  any  calculation. 

93.  From  these  similar  triangles,  are  derived  the  following 
proportions  ;  in  which  R  is  put  for  radius, 

sin  for  sine,  cos  for  cosine, 

tan  for  tangent,  col  for  cotangent, 

sec  for  secant,  cosec  for  cosecant. 

By  comparing  the  triangles  CBG  and  CAD, 

1.  AC  :  BC:  :AD  :  BG,  that  is,  R  :  co^:  :tan  :  sin. 

2.  CG  :  CD:  :BG  :  AD  R  :  secWsin  :  tan. 

3.  CB  :  CA:  :CG  :  CD  cos :  R:  :R  :  sec. 

Therefore  R^  =co5  X  sec. 

By  comparing  the  triangles  CLG  and  CHF, 

4.  CH  :  CL:  :HF  :  LG,  that  i?,  R  :  sin:  :cot  :  COS. 

5.  CG  :  CF::LG  :  HF  R  :  cosecy.cos :  col 

6.  CL  :  CH::CG  :  CF  sin  ?  R::R  :  cosec 

Therefore  R^  =sin  X  cosec* 

By  comparing  the  triangles  CAD  and  CHF, 

7.  CH  :  AD:  :CF  :  CD,  that  is  R  :  tan: : cosec  :  sec, 

8.  CA  :  HF:  :CD  :  CF  R  :  cot: :sec :  cosec. 
7.  AD  :  AC:  :CH  :  HF  tan  :  R:  :R  :  cot. 

Therefore  R^  =  tanXcot. 

It  will  not  be  necessary  for  the  learner  to  commit  these 
proportions  to  memory.  But  he  ought  to  make  himself  so 
familiar  with  the  manner  of  stating  them  from  the  figure,  as 
to  be  able  to  explain  them,  whenever  they  are  referred  to. 


SINES,  TANGENTS,  &c.  53 

'34,  Other  relations  of  the  sine,  tangent,  &c.  may  be  deri- 
Tcd  from  the  proposition,  that  the  square  of  the  hypothenuse 
is  equal  to  the  sum  of  the  squares  of  the  perpendicular  sides. 
(Euc.  47.  1.) 

In  the  right  angled  triangles  CBG,  CAD,  and  CHF, 
(Fig.  3.) 

U     Cg'-=Cb'  +  Bg',  that  is  R2=c05  =  +W7i2  # 

2.  CD'=CA'+AD'  ser.^=R='-{-tan', 

3.  CF"=CH'4-HF'  co^ec2=R2+co<3. 

And,  extracting  the  root  of  both  sides,  (Alg.  296.) 


Yi  =  \/cos'^  -{-sin^  =s^sec^  —tati^  =  \/cosec'^  —cot' 
Hence,  ifR=l,  (Alg.  510.) 


Sine  =  y/l-cos''  Sec  =  \/\+tan^ 

Cos  =  -/ 1  —  sin  2  Cosec=x/l-{-  cot  ^ 

95.  77/6  sine  of  90°  ^ 

The  chord  of  60°  >  are,  in  any  circle,  each   equal  to 


And  the  tangent  of  Ab 
the  radiusj  and  therefore  equal  to  each  other. 

Demonstration* 

1  In  the  quadrant  ACH,  (Fig.  5.)  the  arc  AH  is  90<*.  The 
sine  of  this,  according  to  the  definition,  (Art.  82.)  is  CH,  the 
radius  of  the  circle. 

2.  Let  AS  be  an  arc  of  60°.  Then  the  angle  ACS,  be- 
ing measured  by  this  arc,  will  also  contain  60° ;  (Art.  75.) 
and  the  triangle  ACS  will  be  equilateral.  For  the  sum  of 
the  three  angles  is  equal  to  180°.  (Art.  76.)  From  this 
taking  the  angle  ACS,  which  is  60°,  the  sum  of  the  remain' 
ing  two  is  120°.  But  these  two  are  equal,  because  they  are 
subtended  by  the  equal  sides,  CA  and  CS,  both  radii  of  the 
circle.     Each,  therefore,  is  equal  to  half  120",  that  is  to  60°. 


*  Sines  is  here  put  for  the  square  of  the  sine,   cos 2   for  the  square  ol  the 
cosine  &c. 


o6  TRIGONOMETRY. 

All  the  angles  being  equal,  the  sides  are  equal,  and  therefore 
AS,  the  chord  of  60°,  is  equal  to  CS  the  radius. 

3.  Let  AR  be  an  arc  of  45°.  AD  will  be  its  tangent,  and 
the  angle  ACD  subtended  by  the  arc,  will  contain  45°.  The 
angle  CAD  is  a  right  angle,  because  the  tangent  is,  by  defi- 
nition, perpendicular  to  the  radius  AC.  (Art.  84.)  Subtract- 
ing ACD,  which  is  45°,  from  90°,  (Art.  77.)  the  other  acute 
angle  ADC  will  be  45°  also.  Therefore  the  two  legs  of  the 
triangle  ACD  are  equal,  because  they  are  subtended  by  equal 
angles;  (Euc.  6.  1.)  that  is,  AD  the  tangent  of  45°,  is  equal 
to  AC  the  radius. 

Cor.  The  cotangent  of  45°  is  also  equal  to  radius.  For 
the  complement  of  45°  is  itself  45°.  Thus  HD,  the  cotan- 
gent of  ACD,  (Fig  5.  is  equal  to  AC  the  radius. 

98.  The  sine  of  30°  is  equal  to  half  radius.  For  the  sine 
of  30°  is  equal  to  half  the  chord  of  60°.  (Art.  82.  cor.)  But 
by  the  preceding  article,  the  chord  of  60°  is  equal  to  radius. 
Its  half,  therefore,  which  is  the  sine  of  30°  is  equal  to  lialf 
radius. 

Cor.  1.  The  cosine  of  60°  is  equal  to  half  radius.  For  the 
cosine  of  60°  is  the  sine  of  30°  (Art.  89.) 

Cor.  2.  The  cosine  of  30°  =  |v^3.     For 

Co5230°=R— sm230=l  -\  =  h 

Therefore, 

Cos  30°=v/|  =  i\/3. 

96.  6.  The  sine  of  45°=-77;.       For 

R-=l=5m2  45°-f-co52  45°=2  sm^  45° 
J_ 
72 


Therefore,       Sin  45°  =  v'^  = 


97.  The  chord  of  any  arc  is  a  mean  proportional^  between 
the  diameter  of  the  circle,  and  the  versed  sine  of  the  arc. 

Let  ADB  CFig.  6.)  be  an  arc,  of  which  AB  is  the  chord, 
BF  the  sine,  and  AF  the  versed  sine.  The  angle  ABH  is  a 
right  angle,  (Euc.  31.  3.)  and  the  triangles  ABH  and  ABF 
are  similar.  (Euc.  8.  6.)     Therefore, 

AH  :  AB::AB  :  AF. . 


SINES,  TANGENTS,  kc,  57 

That  is,  the  diameter  is  to  the  chord,  as  the  chord  to  the 
versea  sine. 

In  Fig.  6th,  let  the  arc  AD=a,  and  ADB=2a.  Draw  BF 
perpendicular  to  AH.  This  will  divide  the  right  angled  tri- 
angle ABH  into  two  similar  triangles.  (Euc.  8.  6.)  The  an- 
gles ACD  and  AHB  are  equal.  (Euc.  20.  3.)  Therefore 
the  four  triangles  ACG,  AHB,  FHB,  and  FAB  are  similar; 
and  the  line  BH  is  twice  CG,  because  BH  :  CO  * :  HA  :  Ca! 

The  sides  of  the  four  triangles  are 
AG=5m  cr,        CG =cos  a,  HF  =vers.  sup.  Sa, 

AB  =  2sin  a,      BH  =  2co5  a,         AC=the  radius, 
BF=sin2a,      AF=vers  2a,        AH=the  diameter. 

A  variety  of  proportions  may  be  stated,  between  the  ho- 
mologous sides  of  these  triangles  :     For  instance, 

By  comparing  the  triangles  ACG  and  ABF, 
AC  :  AG:  :AB  :  AF,  that  is,  R  :  sin  a::2s{na  :  vers  2a 
AC  :  CG: :  AB  :  BF  R  :  cos  a:  \'2sin  a  :  sin  2a 

AG  :  CG : :  AF  :  BF  Sin  a  :  cos  a '.  '.vers  2a  t  sin  2a 

Therefore, 
RXvers  2a=2sin^a 
R  X  sin  2a   =2sm  a  X  cos  a 
Sin  aXsin  2a=vers  2a  X  cos  a 

By  comparing  the  triangles  ACG  and  BFH, 
AC  :  CG:  :BH  :  HF,  that  is,  R  :  cos  a:  \2cosa  :  vers,  sup,  2a 
AG  :  CG : : BF  :  HF  Sin  a  l  cos  a :  '.sin  2a  *,  vers,  sup,  2a 

Therefore, 

R  X  vers.  sup.  2a=2cos^  a 
Sin  aXvers.  sup,  2a-=-cos  aXsin  2a 
<Szc.  &tc. 

That  is,  the  product  of  radius  into  the  versed  sine  of  the 
supplement  of  twice  a  given  arc,  is  equal  to  twice  the  square 
of  the  cosine  of  the  arc. 

And  the  product  of  the  sine  of  an  arc,  into  the  versed  sine 
of  the  supplement  of  twice  the  arc,  is  equal  to  the  product  of 
the  cosine  of  the  arc,  into  the  sine  of  twice  the  arc,  <fec.  &c. 

9 


SECTION  II, 


tHE  TRIGONOMETRICAL  TABLES. 


A  qj.  nnO  facilitate  the  operations  in  trigonometry,  the 
sine,  tangent,  secai.t,  <Sic.  have  been  calculated 
for  every  degree  and  minute,  and  in  some  instances,  for  eve- 
ry second,  of  a  quadrant,  and  arrajij^ed  '•:;  ^ibl'^?.  T:.tse 
constitute  what  is  called  the  Trigonometrical  Canon'*  It  is 
not  necessary  to  extend  these  tables  beyond  90°;  because 
the  sines,  tangents,  and  secants,  are  of  the  same  magnitude, 
in  one  of  the  quadrants  of  a  circle,  as  in  the  others.  Thus 
the  sine  of  50°  is  equal  to  that  of  1 50°      (Ai  t.  90.) 

99.  And  in  any  instance,  if  we  have  occasion  for  the  sine, 
tangent,  or  secant  of  an  obtuse  angle,  we  may  obiiiin  ir,  by 
looking  for  its  equal,  the  sine,  tangent,  or  secant  of  the  sup- 
plementary acute  angle. 

100.  The  tables  are  calculated  for  a  circle  whose  radius 
is  supposed  to  be  a  unit.  It  may  be  an  inch,  a  yard,  a  mile, 
or  any  other  denomination  of  length.  But  the  sines^  tau' 
gents^  he,  must  always  be  understood  to  be  of  the  same  de- 
nomination as  the  radius. 

101.  All  the  sines^  except  that  of  90°,  are  less  than  the  ra- 
dius, (Art,  82,  and  Fig.  3.)  and  are  expressed  in  the  tables  by 
decimals. 

Thus  the  sine  of  20°  is  0.34202,         of  60°  is  0.86603, 

of  40°  is  0.64279,         of  89°  is  0.99985,  &c. 

When  the  tables  are  intended  to  be  very  exact,  the  deci- 
mal is  carried  to  a  greater  number  of  places. 

The  tangents  of  all  angles  less  than  45°  are  also  less  than 
radius.  (Art.  95.)  But  the  tangents  of  angles  greater  than 
45°,  are  greater  than  radius,  an'd  are  expressed  by  a  whole 
number  and  a  decimal.     It  is  evident  that  all  the  secants  also 

*  For  the  construction  of  the  Canon,  see  Section  VHl. 


THE  TRIGONOMETRICAL  TABLES.  59 

must  be  greater  than  radius,  as  they  extend  from  the  centre, 
to  a  point  without  the  circle. 

10:2.  The  numbers  in  the  table  here  spoken  of,  are  called 
7iatural  sines,  tangents,  &;c.  They  express  the  lengths  of  the 
several  lines  which  have  been  defined  in  arts.  82,  83,  &c. 
By  means  of  them,  the  angles  and  sides  of  triangles  may  be 
accurately  determined.  But  the  calculations  must  be  made 
by  the  tedious  processes  of  multiplication  and  division.  To 
avoid  this  inconvenience,  another  set  of  tables  has  been  pro- 
vided, in  which  are  inserted  the  logarithms  of  the  natural 
sines,  tangents,  &;c.  By  the  use  of  these,  addition  and  sub- 
traction are  made  to  perform  the  office  of  multiplication  and 
division.  On  this  account,  the  tables  of  logarithmic,  or  as 
they  are  sometimes  called,  artificial  sines,  tangents,  &c.  are 
much  more  valuable,  for  practical  purposes,  than  the  natural 
sines,  &c.  Still  it  must  be  remembered,  that  the  former  are 
derived  from  the  latter.  The  artificial  sine  of  an  angle,  is 
the  logarithm  of  the  natural  sine  of  that  angle.  The  artificial 
tangent  is  the  logarithm  of  the  natural  tangent,  &:c. 

103.  One  circumstance,  however,  is  to  be  attended  to,  in 
comparing  the  two  sets  of  tables.  The  radius  to  which  the 
natural  sines,  &c.  are  calculated,  is  unity.  (Art.  100.)  The 
secants,  and  a  part  of  the  tangents  are,  therefore,  greater  than 
a  unit;  while  the  sines,  and  another  part  of  the  tangents,  are 
/e^5  than  a  unit.  When  the  logarithms  of  these  are  taken, 
some  of  the  indices  will  be  positive,  and  others  negative  ;  (Art. 
9.)  and  the  throsving  of  them  together  in  the  same  table,  if  it 
does  not  lead  to  error,  will  at  least  be  attended  with  incon- 
venience. To  remedy  this,  10  is  added  to  each  of  the  indi- 
ces. (Art.  12.)  They  are  then  all  positive.  Thus  the  natural 
sine  of  20^  is  0.34202.  The  logarithm  of  this  is  T753405, 
But  the  index,  by  the  addition  of  10,  becomes  10—1=9. 
The  logarithmic  sine  in  the  tables  is  therefore  9.53105.^ 

Directions  for  taking  Sines,  Cosines,  &;c,from  the  tables, 

104.  The  cosine,  cotangent,  and  cosecant  of  an  angle,  are 
the  sine,  tangent,  and  secant  of  the  complement  o(  the  an^le. 
(Alt.  89.)  As  the  complement  of  an  angle  is  the  difference 
between  the  angle  and  90^  and  as  45  is  the  half  of  90;  if 
any  given  angle  within  the  quadrant  is  greater  than  45°,  its 

♦Or  the  tables  may  be  supposed  to  be  calculated  to  the  radius  10000000000, 
whose  logarithm  is  10. 


60  tHE  TRIGONOMETRICAL 

complement  is  less  ;  and,  on  the  other  hand,  if  the  angle  is 
less  than  45°,  its  complement  is  greater.  Hence,  every  co- 
sine, cotangent,  and  cosecant  of  an  angle  greater  than  45°. 
has  its  equal,  among  the  sines,  tangents,  and  secants  of  an- 
gles less  than  45°,  and  d.  v. 

Now,  to  bring  the  trigonometrical  tables  within  a  small 
compass,  the  same  column  is  made  to  answer  for  the  sines  of 
a  number  of  angles  abose  45°,  and  for  the  cosines  of  an  equal 
number  below  45°. 

Thus  9.23967  is  the  log.  sine  of  10°,  and  the  cosine  of  80°, 

9.53405  the  sine  of  20°,  and  the  cosine  of  70°,  &c. 

The  tangents  and  secants  are  arranged  in  a  similar  man- 
ner.    Hence, 

105.  To  find  the  Sine,  Cosine,  Tangent,  &lc.  of  any  number 
of  degrees  and  minutes* 

If  the  given  angle  is  less  than  45°,  look  for  the  degrees  at 
the  top  of  the  table,  and  the  minutes  on  the  left  ;  then,  oppo- 
site to  the  minutes,  and  under  the  word  sine  at  the  head  of 
the  column,  will  be  found  the  sine  ;  under  the  word  tangent, 
will  be  found  the  tangent ;  &c. 

The  log.sin  of  43°  25'  is  9.837 1 5  The  tan  of  1 7°  20'  is  9.49430 
ofl7°20'     9.47411  of  8°  46'      9.18812 

The  cos  of  17°  20'  9.97982  Thecotof  17°  20'  10.50570 
of    8°  46'     9.99490  of    8°  46'    10.81188 

The  first  figure  is  the  index  ;  and  the  other  figures  are  the 
decimal  part  of  the  logarithm. 

106.  If  the  given  angle  is  between  45°  and  90°  ;  look  for 
the  degrees  at  the  bottom  of  the  table,  and  the  minutes  on 
the  right  ;  then,  opposite  to  the  minutes,  and  over  the  word 
sine  at  the  foot  of  the  column,  will  be  found  the  sine  ;  over 
the  word  tangent,  will  be  found  the  tangent,  &c. 

Particular  care  must  be  taken,  when  the  angle  is  less  than 
45°,  to  look  for  the  title  of  the  column,  at  the  top,  and  for  the 
minutes,  on  the  left ;  but  when  the  angle  is  between  45°  and 
90°,  to  look  for  the  title  of  the  column,  at  the  bottom  and  for 
the  minutes,  on  the  right. 

The  log.  sine  of  81°  21'  is  9.99503 

The  cosine      of  72°  10'  9.48607 

The  tangent    of  54°  40'  10.14941 

The  cotangent  of  63°  22'  9.70026 


TABLES.  oi 

107.  If  the  given  angle  is  greater  ih^n  90°,  look  for  the 
sine,  tangent,  &:c.  of  its  supplement.     (Art.  98,  99.) 

The  log.  sine  of  96°  44'  is  9.99G99 
The  cosine  of  171°  16'  9.99494 
The  tangent  of  130°  26'  10.06952 
The  cotangent  of  156°  22'    10.35894 

108.  To  find  the  sine,  cosine,  tangent,  <i'C.  of  any  number  of 
degrees,  minutes,  and  seconds. 

In  the  common  tables,  the  sine,  tangent,  «^c.  are  giren  on- 
ly to  every  minute  of  a  degree.*  But  they  may  be  found  to 
seconds,  by  taking  proportional  parts  of  the  difference  of  the 
numbers  as  they  stand  in  the  tables.  For,  within  a  single 
minute,  the  variations  in  the  sine,  tangent,  &;c.  are  nearly  pro* 
portional  to  the  variations  in  the  angle.     Hence, 

To  find  the  sine,  tangent,  &:c.  to  seconds  :  Take  out  the 
number  corresponding  to  the  given  degree  and  minute  ;  and 
also  that  corresponding  to  the  next  greater  minute,  and  find 
their  difference.     Then  state  this  proportion  ; 

As  60,  to  the  given  number  of  seconds  ; 

So  is  the  difference  found,  to  the  correction  for  the  seconds. 

This  correction,  in  the  case  of  sines,  tangents,  and  secants, 
is  to  be  added  to  the  number  answering  to  the  given  degree 
and  minute  ;  but  for  cosines,  cotangents,  and  cosecants,  the 
correction  is  to  be  subtracted 

For,  as  the  sines  increase,  the  cosines  decrease, 

Ex.  1.  What  is  the  logarithmic  sine  of  14°  43'  10"  ? 

The  sine  of  14°  43'  is  9.40490 
of  14°  44'         9.40538 


Difference  43 


Here  it  is  evident,  that  the  sine  of  the  required  angle  ib 
greater  than  that  of  14°  43',  but  less  than  that  of  14°  44'. 
And  as  the  difference   corresponding  to  a  whole   minute  or 

*  In  the  very  valuable  labl«»?  of  Mif^hael  Taylor.  Dip  sines  an^  tangent"?  ar*^ 
•riven  to  even/  '>rrond. 


62  THE  TRIGONOMETRICAL 

60"  is  48  ;  the  difference  for  10"  must  be  a  proportional  pari 
of  48.     That  is, 

60"  :  10":  :48  :  8         the  correction  to  be  ad- 
ded to  the  sine  of  14°  43'. 

Therefore  the  sine  of  14^  43'  10"  is  9.40498 

?.   What  is  the  logarithmic  cosine  of  32°  16'  45"  ? 

The  cosine  of  32°  16'  is  9.92715 
of  32°  17'      9.92707 


Difference  8 


Then  60"  :  45":: 8  :  6  the  correction  to  be  subtracted 
from  the  cosine  of  32°  16'. 

Therefore  the  cosine  of  32°  16'  45"  is  9.92709 

The  tangent  of  24°  15'  18''  is  9.65376 
Thecotangentof  31°  50'  5"  is  10.20700 
The  sine  of  58°  14'  32"         is     9.92956 

The  cosine  of       55°  10'  26"         is     9.75670 
If  the  given  number  of  seconds  be  any  even  part  of  60, 
as  ^,  i,  i,  <Src.  the  correction  may  be  found,  by  taking  a  like 
part  of  the  difference  of  the  numbers  in  the  tables,  without 
stating  a  proportion  in  form. 

109.  To  find  the  degrees  and  minutes  belonging  to  any  given 
sine^  tangent,  (^c. 

This  is  reversing  the  method  of  finding  the  sine,  tangent, 
&c.  (Art.  105,6,  7.) 

Look  in  the  column  of  the  same  name,  for  the  sine,  tan- 
gent, he.  which  is  nearest  to  the  given  one  ;  and  if  the  title 
be  at  the  head  of  the  column,  take  the  degrees  at  the  top  of 
the  table,  and  the  minutes  on  the  left  ;  but  if  the  title  be  at 
xh^foot  of  the  column,  take  the  degrees  at  the  bottom,  and 
the  minutes  on  the  right, 

Ex.  1  .What  is  the  number  of  degrees  and  minutes  belong- 
ing to  the  logartihmic  sine     9.62863  ? 

The  nearest  sine  in  the  tables  is  9.62865.  The  title  of 
sine  is  at  the  head  of  the  column  in  which  these  numbers  are 


iO 


TABLES.  63 

.-  und.  The  degrees  at  the  top  of  the  page  arc  25,  and  the 
minutes  on  the  left  are  10.  The  angle  required  is,  therefore 
25°  10'. 

The  angle  belonging  to 

the  sine  9.87993  is  49^  20'  the  cos  9.97351  is  19^  48 
the  tan  9.97955  43°  39'  the  cotan  9.75791  60°  12' 
the  sec    10.65396      77°  11'       the  cosec  10.49066       18°  51' 

110.  To  find  the  degrees,  minutes,  and  seconds  belonging 
to  any  given  sine,  tangent,  4*c. 

This  is  reversing  the  method  of  finding  the  sine,  tangent, 
tc   to  seconds.  (Art.  108.) 

First  find  the  difference  between  the  sine,  tangent,  Sic. 
next  greater  than  the  given  one,  and  that  which  is  next  less  ; 
then  the  difference  between  this  less  number  and  the  given 
one  ;  Then 

As  the  difference  first  found,  is  to  the  other  difference  ; 

So  are  60  seconds,  to  the  number  of  seconds,  which,  in  the 
case  of  sines,  tangents,  and  secants,  are  to  be  added  to  the 
det^rees  and  minutes  belonging  to  the  least  of  the  two  num- 
bers taken  from  the  tables  ;  but  for  cosines,  cotangents,  and 
cosecants,  are  to  be  subtracted, 

Ex.  1.  What  are  the  degrees,  minutes,  and  seconds,  be- 
longing to  the  logarithmic  sine     9.40498  ? 

Sine  next  greater  14°  44'  9.40538       Given  sine  9.40498 
Next  less         14°  43'  9.40490       Next  less     9.40490 


Difference  48         Difference  8 

Then  48  :  8!  :60''  :  10",         which  added  to   14°   43' 
gives  14°  43'  10"  for  the  answer. 

2  What  is  the  angle  belonging  to  the  cosine  9.09773  ? 

Cosine  next  greater  82°  48'  9.09807     Given  cosine  9.09773 
Next  less  82°  49'  9.09707     Next  less         9.09707 


Differencr  100  Difference  66 


64  THE  TRIGONOMETRICAL 

Then  100  :  66:  :60"  :  40'',     which  subtracted  iVom  82^ 
49',  gives  82°  48'  20"  for  the  answer. 

It  must  be  observed  here,  as  in  all  other  cases,  that  of  the 
two  angles,  the  less  has  the  greater  cosine. 

The  angle  belonging  to 

the  sin  9.20621  is  9°  15'  6"     the  tan  10.43434  is  69^  48'  16" 
the  cos  9.98157    16°  34'  30"  the  cot  10.33554     24°  47'  16" 

Method  of  Supplying  the  Secants  and  Cosecants, 

ill.  In  some  trigonometrical  tables,  the  secants  and  co- 
secants are  not  inserted.  But  they  may  be  easily  obtained 
from  the  sines  and  cosines.     For,  by  art.  93,  proportion  3d, 

cos  X5ec=R2 

That  is,  the  product  of  the  cosine  and  secant,  is  equal  to 
the  square  of  radius.  But,  in  logarithms,  addition  takes  the 
place  of  multiplication  ;  and,  in  the  tables  of  logarithmic  sines, 
tangents,  &ic.  the  radius  is  10.  (Art.  103.)-  Therefore,  in 
these  tables, 

cos-fsec=20.        Or  sec— 20— cos. 
Again,  by  art.  93,  proportion  6, 

sinXcosec=R^. 

Therefore,  in  the  tables, 

sin-{-cosec=20.         Or  cosec=20  — sin.     Hence, 

112.  To  obtain  the  secant^  subtract  the  cosine  from  20; 
and  to  obtain  the  cosecant,  subtract  the  sine  from  20. 

These  subtractions  are  most  easily  performed,  by  taking 
The  right  hand  figure  from  10,  and  the  others  from  9,  as  in 
finding  the  arithmetical  complement  of  a  logarithm  ;  (Art.  55,) 
ob^rving  however,  to  add  10  to  the  index  of  the  secant  or 
cos^ant.  In  fact,  the  secant  is  the  arithmetical  complement 
of  th^  cosine,  with  10  added  to  the  index. 

Foi\the  secant  =20— cos 

And  the  ar.  comp.  of  cos  =10  —  cos.  (Art.  54.) 

So  also  the  cosecant  is  the  arithmetical  complement  of  the 
sine,  with  10  added  to  the  index.  The  tables  of  secants  and 
cosecants  are,  therefore,  of  use.  in  furnishing  the  artihmetical 


TABLES.  65 

complement  of  the  sine  and  cosine,  in  the  following  simple 
manner : 

113.  For  the  arithmetical  complement  of  the  sifie,  subtract 
10  from  the  index  of  the  cosecant;  and  for  the  arithmetical 
complement  of  the  cosine^  subtract  10  from  the  index  of  the 
secant. 

By  this,  we  may  save  the  trouble  of  taking  each  of  the  fig- 
ures from  9. 


10 


SECTION  in. 


SOLUTIONS  OF    RIGHT-ANGLED   TRIANGLES. 


A  114-  T^  ^  triangle,  there  are  six  parts^  three  sides, 
and  three  angles.  In  every  trigonometrical 
calculation,  it  is  necessary  that  some  of  these  should  be  known, 
to  enable  us  to  find  the  others.  The  number  of  parts  which 
must  be  given^  is  three,  one  of  which  must  be  a  side. 

If  only  two  parts  be  given,  they  will  be  either  two  sides,  a 
side  and  an  angle,  or  two  angles ;  neither  of  which  will  limit 
the  triangle  to  a  particular  form  and  size. 

If  two  sides  only  be  given;  they  may  make  any  angle  with 
each  other ;  and  may,  therefore,  be  the  sides  of  a  thousand 
different  triangles.  Thus  the  two  lines  a  and  b  (Fig.  7.)  may 
belong  either  to  the  triangle  ABC,  or  ABC,  or  ABC".  So 
that  it  will  be  impossible,  from  knowing  two  of  the  sides  of  a 
triangle,  to  determine  the  other  parts. 

Or,  if  a  Side  and  an  angle  only  be  given,  the  triangle  will 
be  indeterminate.  Thus,  if  the  side  AB  (Fig.  8.)  and  the 
angle  at  A  be  given  ;  they  may  be  parts  either  of  the  triangle 
ABC,  or  ABC  ,  or  ABC." 

Lastly,  if  two  angles,  or  even  if  a//  the  angles  be  given,  they 
will  not  determine  the  length  of  the  sides.  For  the  triangles 
ABC,  A'B'C,  A"B"C".  (Fig.  9.)  and  a  hundred  others  which 
might  be  drawn,  with  sides  parallel  to  these,  will  all  have  the 
same  angles.  So  that  one  of  the  parts  given  must  always  be 
a  side.  If  this  and  any  other  two  parts,  either  sides  or  angles, 
be  known,  the  other  three  may  be  found,  as  will  be  shown, 
in  this  and  the  following  section. 

115.  Triangles  are  either  right  angled  or  oblique  angled. 
The  calculations  of  the  former  are  the  most  simple,  and  those 
which  we  have  the  most  frequent  occasion  to  make.  A  great 
portion  of  the  problems  in  the  mensuration  of  heights  and  dis- 
tances, in  surveying,  navigation  and  astronomy,  are  solved  by 
rectangular  trigonometry.  Any  triangle  whatever  may  be  di- 
vided into  two  right  angled  triangles,  by  drawing  a  perpen- 
dicular from  one  of  the  angles  to  the  opposite  side. 


RIGHT  ANGLED  TRIANGLES.  6f 

116.  One  of  the  six  parts  in  a  right  angled  triangle,  is  al- 
ways given,  viz,  the  right  angle.  This  is  a  cousiant  quantity  ; 
while  the  other  angles  and  the  sides  are  variahle.  It  is  also 
to  be  observed,  that,  if  one  of  the  acute  angles  is  given,  the 
other  is  known  of  course.  For  one  is  the  complement  of  the 
other.  (Art.  76,  77.)  So  that,  in  a  right  angled  triangle^ 
subtracting  one  of  the  acute  angles  from  90°  gives  the  other. 
There  remain,  then,  on\y  four  parts,  one  of  the  acute  angles, 
and  the  three  sides  to  be  sought  by  calculation.  If  any  two 
of  these  be  given,  with  the  right  angle,  the  others  may  be 
found. 

117.  To  illustrate  the  method  of  calculation,  let  a  case  be 
supposed  in  which  a  light  angled  triangle  CAD  (Fig.  10.) 
has  one  of  its  sides  equal  to  the  radius  to  which  the  trigono- 
metrical tables  are  adapted. 

In  the  first  place,  let  the  haseo{\\\e  triangle  be  equal  to  the 
tabular  radius.  Then,  if  a  circle  be  described,  with  this  ra- 
dius, about  the  angle  C  as  a  centre,  DA  will  be  the  tangent. 
and  DC  the  secant  of  that  angle.  (Art.  84,  85.)  So  that  the 
radius,  the  tangent,  and  the  secant  of  the  angle  at  C,  consti- 
tute the  three  sides  of  the  triangle.  The  tangent,  taken  from 
the  tables  of  natural  sines,  tangents,  &;c.  will  be  the  length  of 
the  perpendicular^  and  the  secant  will  be  the  length  of  the  h^- 
pothenuse.  If  the  tables  used  be  logarithmic,  they  will  give 
the  logarithms  of  the  lengths  of  the  two  sides. 

In  the  same  manner,  any  right  angled  triangle  whatever, 
whose  base  is  equal  to  the  radius  of  the  tables,  will  have  its 
other  two  sides  found  among  the  tangents  and  secants.  Thus, 
if  the  quadrant  AH  (Fig,  11.)  be  divided  into  portions  of  15* 
each;  then,  in  the  triangle 

CAD,  AD  will  be  the  tan,  and  CD  the  sec  of  15°, 
In  CAD',  AD'  wiU  be  the  tan,  and  CD'  the  sec  of  30°, 
In  CAD",  AD"  will  be  the  tan,  and  CD"  the  sec  of  45°, 

[&c. 

118.  In  the  next  place,  let  the  hypothenuse  of  a  right  an- 
gled triangle  CBF  (Fig.  12.)  be  equal  to  the  radius  of  the  ta- 
bles. Then,  if  a  circle  be  described,  with  the  given  radius, 
and  about  the  angle  C  as  a  centre ;  BF  will  be  the  sine^  and 
BC  the  cosine  of  that  angle.  (Art.  82.  89.)  Therefore  the 
sine  of  the  angle  at  C.  taken  from  the  tables,  will  be  the  length 


60  RIGHT  ANGLED 

of  the  perpendicular i  and  the  cosine  will  be  the  length  of  the 
base, 

x4ind  any  right  angled  triangle  whatever,  whose  hopothe- 
nuse  is  equal  to  the  tabular  radius,  will  have  its  other  two 
sides  found  among  the  sines  and  cosines.  Thus  if  the  quad- 
rant AH  (Fig  13.)  be  divided  into  portions  of  15°  each,  in 
the  points  F,  F,'  F",  &:c. ;  then,  in  the  triangle, 

CBF,  FB  will  be  the  sin,  and  CB  the  cos,  of  15°, 
In  CB'F',  FB'  will  be  the  sin,  and  CB'  the  cos,  of  30°, 
In  CB'F",  F"B"  will  be  the  sin,  and  CB"  the  cos,  of  45°, 

[&c. 

119.  By  nnerely  turning  to  the  tables,  then,  we  may  find 
the  parts  of  any  right  angled  triangle  which  has  one  of  its 
sides  equal  to  the  radius  of  the  tables.  But  for  determining 
the  parts  of  triangles  which  have  not  any  of  their  sides  equal 
to  the  tabular  radius,  the  following  proportion  is  used  : 

As  the  radius  of  one  circle, 
To  the  radius  of  any  other  ; 
So  is  a  sine,  tangent,  or  secant,  in  one. 
To  the  sine,  tangent,  or  secant,  of  the  same  number  of 

degrees,  in  the  other. 

In  the  two  concentric  circles  AHM,  ahm,  (Fig.  4,)  the  arcs 
AG  and  ag  contain  the  same  number  of  degrees  (Art.  74  ) 
The  sines  of  these  arcs  are  BG  and  hg,  the  tangents  AD  and 
ad,  and  the  secants  CD  and  cd.  The  four  triangles,  CAD, 
CBG,  Cad,  and  Cbg,  are  similar.  For  each  of  them,  from 
the  nature  of  sines  and  tangents,  contains  one  right  angle ; 
the  angle  at  C  is  common  to  them  all ;  and  the  other  acute 
angle  in  each  is  the  complement  of  that  at  C.  (Art.  77.)  We 
have,  then,  the  following  proportions.  (Euc.  4.  6.) 

1.  CG  :  Cg::BG:%. 

That  is,  one  radius  is  to  the  other,  as  one  sine  to  the  other. 

2.  CA  :  Ca::DA  \da. 

That  is,  one  radius  is  to  the  other,  as  one  tangent  to  the  other. 

3.  CA:Ca::CD:  Cd, 

That  is,  one  radius  is  to  the  other,  as  one  secant  to  the  other. 
Cor.  BG  :  bg\\T>k  :  c/«::CD  :  Cd, 


TRIANGLES.  69 

That  IS,  as  the  sine  in  one  circle,  to  the  sine  in  the  otlier: 
so  is  the  tangent  in  one,  to  the  tangent  in  the  olhei  ;  and  so 
is  the  secant  in  one,  to  the  secant  in  the  other. 

This  is  a  general  principle,  which  may  be  applied  to  most 
trigonometrical  calculations.  If  one  of  the  side.^  of  the  pro- 
posed triangle  be  made  radius,  each  of  the  other  sides  will 
be  the  sine,  tangent,  or  secant,  of  an  arc  described  by  this  ra- 
dius. Proportions  are  then  stated,  between  these  lines,  and 
the  tabular  radius,  sine,  tangent,  &ic. 

120.  A  line  is  said  to  be  made  radius,  when  a  circle  is  de- 
scribed, or  supposed  to  be  described,  whose  semidiameter  is 
equal  to  the  line,  and  whose  centre  is  at  one  end  of  it. 

121.  In  any  right  angled  triangle,  if  the  hypothenuse  be 
made  radius,  one  of  the  legs  will  be  a  sine  of  its  opposite  an- 
gle, and  the  other  leg  a  cosine  of  the  same  angle. 

Thus,  if  to  the  triangle  ABC  (Fig.  14.)  a  circle  be  applied, 
whose  radius  is  AC,  and  whose  centre  is  A,  then  BC  will  be 
the  sine,  and  BA  the  cosine,  of  the  angle  at  A.  (Art.  82,  89.) 

If,  while  the  same  line  is  radius,  the  other  end  C  be  made 
the  centre,  then  BA  will  be  the  sine,  and  BC  the  cosine,  of  the 
angle  at  C. 

122.  If  either  of  the  legs  he  made  radius,  the  other  leg  will 
be  a  TANGENT  of  its  opposite  an^le,  and  the  hypothenuse  will 
be  a  SECANT  of  the  same  angle;  that  is,  of  the  angle  between 
the  secant  and  the  radius. 

Thus,  if  the  base  A.^  (Fig.  15.)  be  made  radius,  the  centre 
being  at  A,  BC  will  be  the  tangent,  and  AC  the  secant,  of 
the  angle  at  A.  (Art.  84,  85.) 

But,  if  the  perpendicular  BC  (Fig.  16.)  be  made  radius, 
with  the  centre  at  C,  then  AB  will  be  the  tangent,  and  AC 
the  secant,  of  the  angle  at  C. 

123.  As  the  side  which  is  the  sine,  tangent,  or  secant  of 
one  of  the  acute  angles,  is  the  cosine,  cotangent,  or  cosecant 
of  the  other;  (Art.  89.)  the  perpendicular  BC  (Fig.  14.)  is 
the  sine  of  the  angle  A,  and  the  cosine  of  the  angle  C  ;  while 
the  base  AB  is  the  sine  of  the  angle  C,  and  the  cosine  of  the 
angle  A. 

If  the  base  is  made  radius,  as  in  Fig.  15,  the  perpendicular 
BC  is  the  tangent  of  the  angle  A,  and  the  cotangent  of  the  an- 
gle C  ;  while  the  hypothenuse  is  the  secant  of  the  angle  A,  and 
the  cosecant  of  the  angle  C. 

If  the  perpendicular  is  made  radius,  as  in  Fig.  16,  the  base 
AB  is  the  tangent  of  the  angle  C,  and  the  cotangent  of  the 


7u  RIGHT  ANGLED 

angle  A;  while  the  hypoihenuse  is  the  secant  of  the  angle  C, 
and  the  co^eca?it  ol'  the  angle  A. 

124.  Whenever  a  right  angled  triangle  is  proposed,  whose 
sides  or  angles  are  required  ;  a  similar  triangle  may  be  formed, 
from  the  sines,  tangents,  &ic.  of  the  tables,  (Art.  117,  118.) 
The  parts  required  are  then  found,  by  stating  proportions  be- 
tween the  similar  sides  of  the  two  triangles.  If  the  triangle 
proposed  be  ABC,  (Fig.  17.J  another  abc  may  be  formed, 
having  the  same  angles  with  the  first,  but  differing  from  it  in 
the  length  of  its  sides,  so  as  to  correspond  with  the  numbers 
in  the  tables.  If  similar  sides  be  made  radius  in  both,  the  re- 
maining similar  sides  will  be  lines  o^  the  same  name ^  that  is, 
if  the  perpendicular  in  one  of  the  triangles  be  a  sine,  the  per- 
pendicular in  the  other  will  be  a  sine  ;  if  the  base  in  one  be  a 
cosine,  the  base  in  the  oiber  will  be  a  cosine,  &;c. 

If  the  hypothenuse  in  each  trian^^le  be  made  radius,  as  in 
Fig.  14,  the  perpendicular  be  will  be  the  tabular  sine  of  the 
angle  at  a ;  and  the  perpendicular  BC  will  be  a  sine  of  the 
equal  angle  A,  in  a  circle  of  which  AC  is  radius. 

If  the  base  in  each  triangle  be  made  radius,  as  in  Fig.  15, 
then  the  perpendicular  be  will  be  the  tabular  tangent  of  the 
angle  at  a  ;  and  BC  will  be  a  tangent  of  the  equal  angle  A,  in 
a  circle  of  which  AB  is  radius,  he. 

125.  From  the  relations  of  the  similar  sides  of  these  trian- 
gles, are  derived  the  two  following  theorems,  which  are  suf- 
ficient for  calculating  the  parts  of  any  right  angled  triangle 
whatever,  when  theVequisite  data  are  furnished.  One  is  used, 
when  a  side  is  to  be  found ;  the  other,  when  an  angle  is  to  be 
found. 

Theorem  I. 

126.  When  a  side  is  required  ; 

As   THE   TABULAR  SINK,  TANGENT,  &tC.  OF 
THE  SAME  NAME  WITH  THE  GIVEN  SIDE, 

To  THE  GIVEN  SIDE; 

So  IS  THE  TABULAR  SINE,  TANGENT,  &;C.  OF 
SAME  NAME   WITH  THE  REQUIRED  SIDE, 

To  THE  REQ,UIRED  SIDE. 

It  will  be  readily  seen,  that  this  is  nothing  more  than  a 
statement,  in  general  terms,  of  the  proportions  between  the 


TRIANGLES.  71 

similar  sides  of  two  triangles,  one  proposed  for  solution,  and 
the  other  formed  from  the  numbers  in  the  tables. 

Thus  if  the  hypothenusfe  be  given,  and  the  base  or  perpen- 
dicular be  requiml ;  then,  in  Fig.  14,  where  ac  is  the  tabular 
radius,  be  the  tabular  sine  of «,  or  its  equal  A,  and  ah  the  tab- 
ular sine  of  C;  (Art.  124.) 

ac  :  AC:  :6c  :  BC,  that  is,  R  :  AC::Sin  A  :  BC. 
ac  :  kCwah  :  AB,  R  :  AC::Sin  C  :  AB. 

In  Fig.  16,  where  ah  is  the  tabular  radius,  ac  the  tabular 
secant  of  A,  and  be  the  tabular  tangent  of  A  ; 

ac  :  AC:  :5c :  BC,  that  is  Sec  A  :  AC:  :Tan  A  :  BC. 
ac  :AC::ah  :  AB,  Sec  A  :  AC::R  :  AB. 

In  Fig.  16,  where  be  is  the  tabular  radius,  ac  the  tabular 
secant  of  C,  and  ab  the  tabular  tangent  of  C  ; 

ac  :  AC ::bc  :  BC,  that  is,  Sec  C  :  AC::R  :BC. 
ac  :  ACy.ab:  AB,  Sec  C  :  AC :  :Tan  C  :  AB. 

Theorem  II. 
127.  When  an  angle  is  required  ; 

As  THE  GIVEN  SIDE  MADE  RADIUS, 
To  THE  TABULAR  RADIUS  ; 
So  15  ANOTHER  GIVEN   SIDE, 

To    THE    TABWLAR    SINE,    TANGENT,  &IC.  OF    THK 
SAME  NAME. 

Thus  if  the  side  made  radius,  and  one  other  side  be  given, 
then,  in  Fig.  14, 

AC  :  ac:  :BC  :  he,  that  is,  AC  :  R:  :BC  :  Sin  A. 
AC  :  ac:: AB  :  ah  AC  :R::AB  :  Sin  C 

In  Fig,  15, 

AB  :  a6::BC  :  be,  that  is,  AB  :  R::BC  :  Tan  A. 
AB  :  abr.AC  :ac  AB  :  R:  :AC  :  Sec  A. 

In  Fig.  16. 

BC  :  he:: AB  :  a6,thatis,  BC  :R::AB  :  Tan  C. 
BC  :  hc::AC  :  er  BC  :  R::AC  :  Sec  C. 


7i  RIGHT  ANGLED 

It  will  be  observed,  that  in  these  theorems,  angles  are  not 
introduced,  though  they  are  among  the  quantities  which  are 
either  s'ven  or  required,  in  the  calculation  of  triangles.  But 
the  tabular  sines,  tangents,  &£c.  may  be  considered  the  repre- 
sentatives of  angles,  as  one  may  be  found  from  the  other,  by 
merely  turning  to  the  tables. 

128.  In  the  theorem  for  finding  a  side,  the  first  term  of 
the  proportion  is  a  tabular  number.  But,  in  the  theorem  for 
finding  an  angle,  the  fir^t  term  is  a  side.  Hence,  in  applying 
the  proportions  to  particular  cases,  this  rule  is  to  be  observed, 

To  find  a  side,  begin  xvith  a  tabular  number, 
To  find  an  ANfJLE,  begin  with  a  side. 

Radius  is  to  be  reckoned  among  the  tabular  numbers. 

129.  In  the  theorem  for  finding  an  angle,  the  first  term  is  a 
side  made  radius.  As  in  every  proportion,  the  three  first 
terms  must  be  given,  to  enable  us  to  find  the  fourth,  it  is  evi- 
dent, that  where  this  theorem  is  applied,  the  side  made  radi- 
us must  be  a  given  one.  But,  in  the  theorem  for  finding  a 
side,  it  is  not  necessary  that  either  of  the  terms  should  be  ra- 
dius.    Hence, 

130.  To  find  a  side,  any  side  may  be  made  radius. 

To  find  an  angle,  a  given  side  must  be  made  radius. 

It  will  generally  be  expedient,  in  both  cases,  to  make  radi- 
us one  of  the  terms  in  the  proportion ;  because,  in  the  tables 
of  natural  sines,  tangents,  iic.  radius  is  1,  and  in  the  logarith- 
mic tables  it  is  10.  (Art.  103.) 

131.  The  proportions  in  Trigonometry  are  of  the  same 
nature  as  other  simple  proportions.  The  fourth  term  is  found, 
therefore,  as  in  the  Rule  of  Three  in  Arithmetic,  by  multiply- 
ing together  the  second  and  third  terms,  and  dividing  their  pro- 
duct by  the  first  term.  This  is  the  mode  of  calculation,  when 
the  tables  o^  natural  sines,  tangents,  Sic.  are  used.  But  the 
operation  by  logarithms  is  so  much  more  expeditious,  that  it 
has  almost  eniirfcly  superseded  the  other  method.  In  loga- 
rithmic calcut^iioas,  addition  takes  the  place  of  multiplica- 
tion ;  and  subtraction  the  place  of  division. 

The  logarithms  expressing  the  lengths  of  the  sides  of  a  tri- 
an^!:  f^re  to  be  taken  from  the  tables  of  common  logarithms. 
Tiv:  logarithms  of  the  sines,  tangents,  he.  are  found  in  the  ta- 
bles of  artificial  sines,  &c.     The  calculation  is  then  made  by 


TRIANGLES,  73 

adding  the  f^econd  and  third  terms^  and  subtracting   the  first, 
(Art.  b%) 

132.  The  logarithmic  radius  10,  or,  as  it  is  written  in  the 
tables,  10.00000,  is  so  easily  added  and  tJubtracted,  that  the 
three  terms  of  which  it  is  one,  may  be  considered  as,  in  ef- 
fect, reduced  to  two.     Thus,  if  Ihf*  tabular  radius  is   in   the 

first  term,  we  have  only  to  add  the  other  two  terms,  and 
then  take  10  from  the  index  ;  for  this  is  subtracting  the  first 
term.  If  radius  occurs  in  the  second  term,  the  first  is  to  be 
subtracted  from  the  third,  after  its  index  is  increased  by  10. 
In  the  same  manner,  if  radius  is  in  the  third  term,  the  first  is 
to  be  subtracted  from  the  second. 

133.  Every  species  of  right  angled  triangles  maybe  be  sol- 
ved upon  the  principle,  that  the  sides  of  similar  triangles  are 
proportional,  according  to  the  two  theorems  mentioned  above. 
There  will  be  some  advantages,  however,  in  giving  the  exam- 
ples in  distinct  classes. 

There  must  be  given,  in  a  right  angled  triangle,  tzoo  of  the 
parts,  besides  the  right  angle.  (Art.  1 16.)     These  may  be  ; 

1.  The  hypothenuse  and  an  angle  ;  or 

2.  The  hypothenuse  and  a  leg  ;  or 

3.  A  leg  and  an  angle  ;  or 

4.  The  two  less. 


Case  I. 

io>.    r-         S  The  hypothenuse,  }  .    r^  i    S  The  base  and 
134.  Given  <  *    j  ..  i^  >  to  fand    <  ^  i;^  i 

I  And  an  angle  :         ^  (  Perpendicular. 

Ex.  1.  If  the  hypothenuse  AC  (Fig.  17.*)  be  45  miles,  and 
the  angle  at  A  32^  20',  what  is  the  length  of  the  base  AB, 
and  the  perpendicular  BC  ? 

In  this  case,  as  sides  only  are  required,  ani/  side  may  be 
made  radius.  (Art.  130.) 

If  the  hypothenuse  be  made  radius,  as  in  Fig.  14,  BC  will 
be  the  sine  of  A,  and  AB  the  sine  of  C,  or  the  cosine  of  A. 
(Art.  121.)  And  if  abc  be  a  similar  triangle,  whose  hypoth- 
enuse is  equal  to  the  tabular  radius,  be  will  be  the  tabular 
sine  of  A,  and  ab  the  tabular  sine  of  C.  (Art.  124.) 

*  The  parts  which  are  given  are  distinguished  by  a  mark  across  the  line,  or 
at  the  opening  of  the  angle,  and  the  parts  required^  by  a  cipher, 

1! 


74  RIGHT  ANGLED 

To  find  the  perpendicular,  then,  by  Theorem  I,  wc  have 
this  proportion  ; 

ac :  AC: :6c :  BC. 
Or   R  :  AC::SinA  :  BC. 

Whenever  the  terms  Radius,  Sine,  Tangent,  &c.  occur  in 
a  proportion  like  this,  the  /a^w/ar  Radius,  &c.  is  to  be  under- 
stood, as  in  arts.  126,  127. 

The  numerical  calculation,  to  find  the  length  of  BC,  may 
be  made,  either  by  natural  sines,  or  by  logarithms.  See  art. 
131. 

By  natural  Sines. 

\  :  45::0.53484  :  24.068=BC. 

Computation  by  Logarithms, 

As  Radius  10.00000 

To  the  hypothenuse         45  1.65321 

So  is  the  Sine  of  A  32°  20'  9.72823 


To  the  perpendicular     24.068  1.38144 


Here,  the  logarithms  of  the  second  and  third  terms  are  ad- 
ded, and  from  the  sum,  the  first  term  10  is  subtracted.  (Art. 
132.)  The  remainder  is  the  logarithm  of  24.068  =  BC. 

Subtracting  the  angle  at  A  from  90°,  we  have  the  angle  at 
C==57°  40'.  (Art.  1 16.)     Then,  to  find  the  base  AB  ; 

ac  I  AC:  °.ab  t  AB 
Or    R  :  AC::SinC  :  AB=38.023. 

Both  the  sides  required  are  now  found,  by  making  the  hy- 
pothenuse radius.  The  results  here  obtained  may  be  verifi- 
ed, by  making  either  of  the  other  sides  radius. 

If  the  base  be  made  radius,  as  in  Fig.  15,  the  perpendicu- 
lar will  be  the  tangent,  and  the  hypothenuse  the  stcant  of  the 
angle  at  A.  (Art.  122.)     Then, 

Sec  A  :  AC::R:  AB 
R:  AB::Tan  A  :  BC 


TRIANGLES.  76 

By  making  the  arithmetical  calculations,  in  these  two  pro- 
portions, the  values  of  AB  and  BC  will  be  found  the  same  as 
before. 

U  the  perpendicular  be  made  radius,  as  in  Fig.  iO,  AB  will 
be  tbe  tangent,  and  AC  the  secant  of  the  angle  at  C.     Then, 

Sec  C  :  AC::R  :  BC 
R  :  BC::TanC  :  AB 

Ex.  2.  If  the  hypothenuse  of  a  right  angled  triangle  be 
250  rods,  and  the  angle  at  the  base  46^  30'  ;  what  is  the 
length  of  the  base  and  perpendicular  ? 

Ans.  The  base  is  172.1  rods,  and  the  perpendic.  181.35. 

Case  II. 

lOK    n^.r^^   ^  The  hypothenuse,  >  ,     ^    ,    (  The  aneles  and 
135.  Given   ^  ^^^  ^^^  ^^^  ^  ,«  find    ^  ^^^  ^^^^^  1^^^ 

Ex.  1.  If  the  hypothenuse  (Fig;.  18.)  be  35  leagues,  and 
the  base  26  ;  what  is  the  length  of  the  perpendicular,  and  the 
quantity  of  each  of  the  acute  angles  ? 

To  find  the  angles  it  is  necessary  that  one  of  the  given  sides 
be  made  radius.  (Art.  130.) 

If  the  hypothenuse  he  radius,  the  hsise  and  perpendicular 
will  be  sines  of  their  opposite  angles.     Then, 

AC  :  R::AB  :  Sin  C=47°  58'i 

And  to  find  the  perpendicular  by  Theorem  I ; 

R  :  AC::Sin  A  :  BC=23.43 

If  the  base  be  radius,  the  perpendicular  will  be  tangent, 
and  the  hypothenuse  secant  of  the  angle  at  A.     Then, 

AB  :  R::AC  :  Sec  A 
R  :  AB::TanA  :  BC 

In  this  example,  where  the  hypothenuse  and  base  arc  giv- 
en, the  angles  can  not  be  found  by  making  the  perpendicular 
radius.  For  to  find  an  angle,  a  gzr^/i  side  must  be  made  ra- 
dius. (Art.  130.) 


76  RIGHT  ANGLED 

136.  Ex.  2.  If  the  hypothenuse  (Fig.iy.)  be  64  milesj. 
and  the  perpendicular  48  miles,  what  are  the  angles,  and  the 
base  ? 

Making  the  hypothenust  radius. 

AC  :  R::BC  :  Sin  A 
R:  AC::SinC  :  AB 

The  numerical  calculation  will  give  A=62°  44',  and  AB 
=24.74. 

Making  the  perpendicular  radius, 

BC  :R::AC  :  Sec  C 
R:  BC::TanC  :  AB 

The  angles  can  not  be  found  by  making  the  base  radius, 
when  its  length  is  not  given. 


Case  III. 

137.  Given    11^^""^^^    ?to  find   f  1^",^^"^"' 
5  A"d  one  leg   )  (.  And  the  other  leg. 

Ex.  1.  If  the  base  (Fig.  20,)  be  60,  and  the  angle  at  the 
base  47°  12',  what  is  the  length  of  the  hypothenuse  and  the 
perpendicular  ? 

In  this  case,  as  sides  only  are  required,  any  side  may  be  ra- 
dius. 

Making  the  hypothenuse  radius. 

SinC  :  AB::R  :  AC  =  88.31 
R  :  AC::Sin  A  :  BC=64.8 

Making  the  base  radius. 

R  :  AB::SecA  :  AC 
R  :  AB::Tan  A  :  BC 

Making  the  perpendicular  radius^ 

TanC  :  AB::R:BC 
R:BC::SecC  :  AC 


TRIANGLES.  77 

138.  Ex.  2.  If  the  perpendicular  (Fig.  21.)  be  74,  and 
the  angle  C  61°  27',  what  is  the  length  of  the  base  and  the 
hjpothenuse  ? 

Making  the  hypothenuse  radius. 

Sin  A  :  BC::R  :  AC 
R  :  AC::SinC  :  AR 

Making  the  base  radius. 

Tan  A  :  BC::R:  AB 
R:  AB::SecA  :  AC 

Making  ihe  perpendicular  radius. 

R  :  BC::SecC  :  AC 
R  :  BC::TanC  :  AB 

The  hypothenuse  is  154.83  and  the  base  136. 
Case  IV. 


139.  Given 


(  The  base,  and  ?  x    ^   j    J  Tlie  hypothenuse, 
I  Perpendicular  5  ^^  """^    {  And  the  angles. 

Ex.  1.  If  the  base  (Fig.  22,  be  284,  and  the  perpendicular 
192,  what  are  the  angles,  and  the  hypothenuse  ? 

In  this  case,  one  of  the  legs  nnust  be  made  radius,  to  find 
an  angle;  because  the  h)'pothenuse  is  not  given. 

Making  the  base  radius. 

AB:  R:.:BC  :  Tan  A  =  34M' 
R  :  AB:  :Sec  A  :  AC  =  342.84 

Making  the  perpendicular  radius. 

BC  :  R::AB:  Tan  C 
R:BC::SecC  :  AC 

Ex.  2.  If  the  base  be  G40,  and  the  perpendicular  480, 
what  are  the  angles  and  hypothenuse  ? 

Ans.  The  hypothenuse  is  800,  and  the  angle  at  the  base 
36°  52'  12". 


78  KIGHT  ANGLED 

Examples  for  practice, 

1.  Given  the  hypothenuse   68,  and  the  angle  at  the  base 
39°  17'  ;  to  find  the  base  and  perpendicular. 

2.  Given  the  hypothenuse   850,  and  the  base   594,  to  find 
the  angles,  and  the  perpendicular. 

3.  Given  the  hypothej-ase  78  and  perpendicular  57,  to  find 
the  base,  and  the  angles. 

4.  Given  the  base  723,  and  the  angle  at  the  base  64°   18', 
to  find  the  hypothenuse  and  perpendicular. 

5.  Given  the  perpendicular  632,  and  the  angle  at  the  base 
81°  36',  to  find  the  hypothenuse  and  the  base. 

6.  Given  the   base  32,  and   the  perpendicular  24,  to   find 
the  hypothenuse,  and  the  angles. 

140.  The  preceding  solutions  are  all  effected,  by  means  of 
the  tabular  sines,  tangents,  and  secants.  But,  when  any  two 
sides  of  a  right  angled  triangle  are  given,  the  third  side  may 
be  found,  without  the  aid  of  the  trigonometrical  tables,  by 
the  proposition,  that  the  square  of  the  hypothenuse  is  equal  to 
the  sum  of  the  squares  of  the  two  perpendicular  sides  (Euc. 
47.  1.) 

If  the  legs  be  given,  extracting  the  square  root  of  the  sum 
of  their  squares,  will  give  the  hypothenuse.  Or,  if  the  hy- 
pothenuse and  one  leg  be  given,  extracting  the  square  root  of 
the  difference  of  the  squares,  will  give  the  other  leg. 

Let  A=the  hypothenuse   ^ 

^=the  perpendicular  >  of  a  right  angled  triangle. 
6=the  base  S 


Then  h^=^b'--\-p\  or  (Alg.  296.)  h=^b^+p^ 

By  trans.       b^=h^-p'',  or  b=\/h^  -p"" 

And  p^-=h^-b\  or  p  =  ^h^-b^ 

Ex.  1.  If  the  base  is  32,  and   the  perpendicular  24,  what 
is  the  hypothenuse  ?  Ans.  40. 

2.  If  the  hypothenuse  is  100,  and  the  base  80,  what  is  the 
perpendicular?  Ans.  60. 

3.  If  the  hypothenuse  is  300,  and  the  perpendicular  220, 
what  is  the  base  ? 


Ans.  300    —220=4160,  the  root  of  which  is  204  nearly. 


TRIANGLES.  79 

141.  It  is  generally  most  convenient  to  find  the  difference 
of  the  squares  by  logarithms.  But  this  is  not  to  be  done  by 
subtraction.  For  subtraction^  in  logarithms,  performs  the  of- 
fice of  division.  (Art.  41.)  If  we  subtract  the  logarithm  of 
b^  from  the  logarithm  of  /i^,  we  shall  have  the  logarithm, 
not  of  the  difference  of  the  squares,  but  of  their  quotient. 
There  is,  however,  an  indirect,  though  very  simple  method, 
by  which  the  difference  of  the  squares  may  be  obtained  by 
logarithms.  It  depends  on  the  principle,  that  the  dfference 
of  the  squares  of  two  quantities  is  equal  to  the  product  of  tlw. 
sum  and  difference  of  the  quantities.  (Alg.  235.)     Thus 

t      h^~-b^-={h-\-b)X{h—b) 

as  will  be  seen  at  once,  by  performing  the  multiplication. 
The  two  factors  may  be  multiplied  by  adding  their  loga- 
rithms.    Hence, 

142.  To  obtain  the  difference  of  the  squares  of  tzoo  quanti- 
ties, add  the  logarithm  of  the  sum  of  the  quantities,  to  the  log- 
arithm of  their  dfference.  After  the  logarithm  of  the  differ- 
ence of  the  squares  is  found ;  the  square  root  of  this  differ- 
ence is  obtained,  by  dividing  the  logarithm  by  2.  (Art.  47.) 

Ex,  1.  If  the  hypothenuse  be  75  inches,  and  the  base  45, 
what  is  the  length  of  the  perpendicular? 

Sum  of  the  given  sides  120         log.   2,07918 

Difference  of  do.  30  1.47712 


Dividing  by         2)3.55630 
Side  required  60  1.77015 

2.    if  the  hypothenuse  is  135,  and  the  perpendicular  108- 
what  is  the  length  of  the  base  ?  Ans.  81. 


SECTION  IV. 


SOLUTIONS  OF  OBLIQUE  ANGLED  TRIANGLES. 


A        14^     T^HE  sides  and  angles  of  oblique  angled  trian- 
^^'        *     -*-   gles  may  be  calculated  by  the  following  the- 
orems. 

Theorem  I. 

In  any  plane  triangle,  the  sines  of  the   angles  are  as 

THEIR  OPPOSITE  SIDES. 

Let  the  angles  be  denoted  by  the  letters  A,  B,  C,  and  their 
opposite  sides  by  cr,  6,  c,  as  in  Fig.  23  and  24.  From  one  of 
the  angles,  let  the  line  p  be  drawn  perpendicular  to  the  op- 
posite side.  This  will  fail  either  within  or  without  the  tri- 
angle. 

1.  Let  it  fall  within  as  in  Fig.  23.  Then,  in  the  right  an- 
gled triangles  ACD  and  BCD,  according  to  art.  126, 

R  :  6:  :Sin  A  :  p 
R  :  a::SinB  :  p 

Here,  the  two  extremes  are  the  same  in  both  proportions. 
The  other  four  terms  are,  therefore,  reciprocally  proportion- 
al: (Alg.  387.*)  that  is, 

a  :  i::Sin  A  :  SinB. 

2.  Let  the  perpendicular  p  fall  ivithoiit  the  triangle,  as  in 
Fig.  24.     Then,  in  the  right  angled  triangles  ACD  and  BCD  : 

R  :  6:  :Sin  Alp 
R  :  a:  :Sin  B  :  p 

Therefore  as  before, 
a  :  6: : Sin  A  J  SinB. 

*  Euclid  23.  5 


OBLIQUE  ANGLED  TRIANGLES  81 

Sin  A  is  here  put  both  for  the  sine  of  DAC,  and  for  that 
of  BAG.  For,  as  one  of  these  angles  is  the  supplement  of 
the  other,  they  have  the  same  sine.  (Art.  90.) 

The  sines  which  are  mentioned  here,  and  which  are  used 
in  calculation,  are  tabular  sines.  But  the  proportion  will  be 
the  same,  if  the  sines  be  adapted  to  any  other  radius.  (Art. 
119.) 

Theorem  !!• 
3  44.   In  a  plane  triangle, 

As  THE  SUM  OF  ANY  TWO  OF  THE  SIDES, 

To  THEIR  DIFFERENCE  ; 

So    IS    THE    TANGENT    OF  HALF  THE  SUM  OF  THE 

OPPOSITE  ANGLES  ; 
To  THE  TANGENT  OF  HALF  THEIR  DIFFERENCE. 

Thus  the  sum  of  AB  and  AC  (Fig.  25)  is  to  their  differ- 
ence ;  as  the  tangent  of  half  the  sum  of  the  angles  ACB  and 
ABC,  to  the  tangent  of  half  their  difference. 

Demonstration. 

Extend  CA  to  G,  making  AG  equal  to  AB  ;  then  CG  is 
Uie  sum  of  the  two  sides  AB  and  AC.  On  AB,  set  off  AD 
equal  to  AC  ;  then  BD  is  the  difference  of  the  sides  ABand  AC. 

The  sum  of  the  two  angles  ACB  and  ABC,  is  equal  to  the 
sum  of  ACD  and  ADC ;  because  each  of  these  sums  is  the 
supplement  of  CAD.  (Art.  79.)  But,  as  AC  =  AD  by  con- 
struction, the  angle  ADC  =  ACD.  (Euc.  5.  1.)  Therefore 
ACD  is  half  the  sum  of  ACB  and  ABC.  As  AB=AG,  the 
angle  AGB  =  ABG  or  DBE.  Also  GCE  or  ACD=ADC=: 
BDE.  (Euc.  15.  1.)  Therefore,  in  the  triangles  GCE  and 
DBE,  the  two  remaining  angles  DEB  and  CEG  are  equal  : 
(Art.  79.)  So  that  CE  is  perpendicular  to  BG.  (Euc.  Def. 
10.  1.)  If  then  CE  is  made  radius,  GE  is  the  tangent  of  GCE, 
(Art,  84.)  that  is,  the  tangent  of  half  the  sum  of  the  angles  op- 
posite to  AB  and  AC. 

If  from  the  greater  of  the  (wo  angles  ACB  and  ABC,  there 
be  taken  ACD  their  half  sum  ;  the  remaining  angle  ECB 
will  be  their  half  difference.  ( Alg.  34 1 .)  The  tangent  of  this 
angle,  CE  being  radius,  is  EB,  that  is,  the  tangent  of  half 
the.  dffcrence  oj  the  angles  opposite  to  AB  and  AC.  We  have 
then, 

I? 


Sii  OBLiqUE  ANGLED 

CG=tiie  iiim  of  the  sides  AB  and  AC; 

T)B=t]ieir  difference  ; 

GE=the  tangent  of  half  the  sum  of  the  opposite  angles  5 

f'iB=the  tangentof  half  their  difference. 

But,  by  similar  triangles, 
CG  :DB::GE  :  EB.  Q.  E.  D. 

Theorem  III. 

145.  If  upon  the  longest  side  of  a  triangle,  a  perpcndicu] 
Jar  be  drav/n  from  the  opposite  angle ; 

As  THE  LONGEST  SIDE, 
To  THE  SUM  OF  THE  TWO  OTHERS  ; 
So  IS  THE  DIFFERENCE  OF  THE  LATTER. 
To  THE  DIFFERENCE  OF  THE  SEGMENTS  MADE  BY 
THE  PERPENDICULAR. 

in  the  triangle  ABC,  (Fig.  26.)  if  a  perpendicular  be  drawn 
from  C  upon  AB; 

AB  :  CB+CA::CB -CA  :  BP-PA.* 

Demonstration* 

Describe  a  circle  on  the  centre  C,  and  with  the  radius  BC. 
Through  A  and  C,  draw  the  diameter  LD,  and  extend  BA 
10  H.     Then  by  Euc.  35,  3, 

ABXAH=ALXAD 

Therefore, 

AB  :  AD::AL  :  AH 

J^ut  AD=CD4-CA=CB-FCA 
And  AL=:CL-CA=CB-CA 
AndAII  =  HP-PA  =  BP-PA  (Euc.  3.  3.) 

If  then,  for  the  three  last  terms  in  the  proportion,  we  sub* 
atitute  their  equals,  we  have, 

AB  :  CB+CA::CB-CA  :  PB-PA 

146.  It  is  to  be  observed,  that  the  greater  segment  is  next 
the  areater  side.     If  BC  is  greater  than  AC,  (Fig.  26.)  PB  is 

*  See  note  F. 


TRIANGLES.  b:^ 

greater  than  AP.    With  the  radius  AC,  describe  the  arc  AN. 
The  segment  NP=AP.  (Euc.  3.  3.)  But  BP  is  greater  than 

NP. 

147.  The  two  segments  are  to  each  other,  as  the  tangents 
of  the  opposite  angles,  or  the  cotangents  of  the  adjacent  an- 
gles. For,  in  the  right  angled  triangles  ACP  and  BCP. 
(Fig.  26.)  if  CP  be  made  radius,  (Art.  126.) 

R:PC::Tan  ACP:  AP 
R:PC::TanBCP  :  BP 

Therefore,  by  equality  of  ratios,  (Alg.  384.*) 
Tan  ACP  :  AP::TanBCP  :  BP 

That  is,  the  segments  are  as  the  tangents  of  the  opposite 
angles.  And  the  tangents  of  these  are  the  cotangents  of  the 
angles  A  and  B.   (Art.  89.) 

Cor.  The  greater  segment  is  opposite  to  the  greater  an- 
gle. And  of  the  angles  at  the  base,  the  less  is  next  the 
greater  side.  If  BP  is  greater  than  AP,  the  angle  BCP  i^ 
greater  than  ACP  :  and  B  is  less  than  A.  (Art.  77.) 


148.  To  enable  us  to  find  the  sides  and  angles  of  an  ob- 
lique angled  triangle,  three  of  them  must  be  given.  (Art.  1 1 4.) 

These   may  be,  cither 

1.  Two  angles  and  a  side,  or 

2.  Two  sides  and  an  angle  opposite  one  of  them,  oi- 

3.  Two  sides  and  the  included  angle,  or 

4.  The  three  sides. 

The  two  first  of  these  cases  are  solved  by  theorem  1,  (An. 
143.)  the  third  by  theorem  II,  (Art.  144.)  and  the  fourth  bv 
theorem  III,  (Art.  145.) 

149.  In  making  the  calculations,  it  must  be  kept  in  mind, 
that  the  greater  side  is  always  opposite  to  the  greater  angle, 
(Euc.  18,  19.  I.)  that  there  can  be  only  one  obtuse  angle  in  a 
triangle,  (Art.  76.)  and  therefore,  that  the  angles  opposite  to 
the  two  lea«t  sides  must  be  acute. 


oBJJQUE  ANGLKI 
Case  I. 


150.  Given. 
Two  ancrl<^?-^rind  f  ,     r    i  S  -t^'c  remaining  angle,  and 
A  side.  5        ^       (  The  other  two  sides. 

The  third  angle  is  found,  by  merely  subtracting  the  sura  of 
the  two  which  are  given  from  180^  (Art.  79  ) 

The  sides  are  found,  by  stating,  according  to  theorem  I, 
the  following  proportion  ; 

As  tiie  sine  of  the  angle  opposite  the  given  side, 
To  the  length  of  the  given  side  ; 
So  is  the  sine  of  the  angle  opposite  the  required  side, 
To  the  length  of  the  required  side. 

As  a  side  is  to  be  found,  it  is  necessary  to  begin  with  a  tab' 
ular  number, 

Ex.  1.  In  the  triangle  ABC  (Fig.  27.)  the  side  h  is  given 
32  rods,  the  angle  A  56°  20',  and  the  angle  C  49°  10',  to  find 
the  angle  B,  and  the  sides  a  and  c. 

The  sum  of  the  two  given  angles  56°  20'-}-49°  10'=  105* 
30';  which  subtracted  from  180°,  leaves  74°  30'  the  angle  B. 
Then, 

c-    -D     1  . .    (  Sin  A  :  a 
.smB:6..    ^  gin  C  :  ,: 

Calculation  by  logarithms. 

As  the  Sine  of  B  74°  30'         a.  c.     0.01609 

To  the  side  ^^  32  1.50515 

So  is  the  Sine  of  A      56°  20'  9.92027 


To  the  side  a 

27.64 

74°  30^      «. 

32 

49°  10* 

25.13 

1.44151 

As  the  Sine  of  B 

To  the  side  b 

So  is  the  Sine  of  C 

c.     0.01609 
1.50515 
9.87887 

To  the  side  c 

1.40011 

TRIANGLES.  06 

The  arithmetical  compUment  used  in  the  tirst  term  here, 
maybe  found,  in  the  usual  way,  or  by  taking  out  the  cosecant 
of  the  given  angle,  and  rejecting  10  from  the  index.  (Art.  113.) 

Ex.  2.  Given  the  side  h  71.  the  angle  A  107°  6'  and  the 
angle  C  27°  40  ;  to  find  the  angle  B,  and  the  sides  a  and  c. 
The  angle  B  is  45°  14'.     Then 


Sin  B  :  h 


(Sin 
(Sin 


A  :  a=95.58 
C  :c  =  46.43 


When  one  of  the  given  angle?  is  obtuse,  as  in  this  example, 
the  sine  of  its  supplement  is  to  be  taken  from  the  tables.  (Art. 
99.) 

Case  11. 

151.  Given 
Two  sides,  and        ^  t    fi    1    5  "^^^  remaining  side,  and 
An  opposite  angle,  )        ^      \  The  other  two  angles. 

One  of  the  required  angles  is  found,  by  beginning  with  a 
-ide,  and,  according  to  theorem  I,  stating  the  proportion. 

As  the  side  opposite  the  given  angle, 
To  the  sine  of  that  angle  ; 
So  is  the  side  opposite  the  required  angle, 
To  the  sine  of  that  angle. 

The  third  angle  is  found,  by  subtracting  the  sum  of  the  oth- 
er two  from  180"^  ;  and  the  remaining  side  is  found,  by  the 
proportion  in  the  preceding  article. 

152.  In  this  second  case,  if  the  side  opposite  to  the  given 
angle  be  shorter  than  the  other  given  side,  the  solution  will 
be  ambiguous.  Two  different  triangles  may  be  formed,  each 
-of  which  will  satisfy  the  conditions  of  the  problem. 

Let  the  side  6,  (Fig.  28.)  the  angle  A,  and  the  length  of 
the  side  opposite  this  angle  be  given.  With  the  latter  for  ra- 
dius, (if  it  be  shorter  than  6,)  describe  an  arc,  cutting  the  line 
AH  in  the  points  B  and  B'.  The  lines  BC  and  B'C  will  be 
equal.  So  that,  with  the  same  data,  there  may  be  formed 
two  different  triangles.  ABC  and  AB'C. 


6u  UBLiqUK  ANGLED 

There  will  be  the  same  ambiguity  in  the  numerical  calcu- 
lation. The  answer  found  by  the  proportion  will  be  the  sine 
of  an  an^le.  But  this  may  be  the  sine,  cither  of  the  acute 
angle  ABC,  or  of  the  obtuse  angle  ABC.  For,  BC  being 
equal  to  B'C,  the  angle  CB'B  is  equal  to  CBB'.  Therefore 
ABC,  which  is  the  supplement  of  CBB'  is  also  thesupplement 
of  CB'B.  But  the  sine  of  an  angle  is  the  same,  as  the  sine 
of  its  supplement.  (Art.  90.)  The  result  of  the  calculation 
will,  therefore  be  ambiguous.  In  practice  however,  there 
will  generally  be  some  circumstances  which  will  determine 
whether  the  angle  required  is  acute  or  obtuse. 

If  the  side  opposite  the  given  angle  be  longer  than  the 
other  given  side,  the  angle  which  is  subtended  by  the  latter, 
will  necessarily  be  acute.  For  there  can  be  but  one  obtuse 
angle  in  a  triangle,  and  this  is  always  subtended  by  the  long- 
est side.  (Art.  149.) 

If  the  given  angle  be  obtuse,  the  other  two  will,  of  course, 
be  acute.  There  can,  therefore,  be  no  ambiguity  in  the  so- 
lution. 

Ex.  1.  Given  the  angle  A,  (Fig.  28.)  35°  20',  the  oppo- 
site side  a  50,  and  the  side  6  79  ;  to  tind  the  remaining  side, 
and  the  other  two  angles. 

To  find  the  angle  opposite  to  b,  (Art.  151.) 

c  :  Sin  A:  :6  :  Sin  B 

The  calculation  here  gives  the  acute  angle  AB'C  54°3'  50''. 
and  the  obtuse  angle  ABC  125°  56'  10''.  If  the  latter  be 
added  to  the  angle  at  A  35°  20',  the  sum  will  be  1 6 1  °  1 6'  1 0", 
the  supplement  of  which  18**  43'  50"  is  the  angle  ACB. 
Then  in  the  triangle  ABC,  to  find  the  side  c=AB, 

Sin  A  :  a:  tSin  C  :  c=27.7G 

If  the  acute  angle  AB'C  54°  3'  50'  be  added  to  the  angle 
at  A  3o°  20',  the  sum  will  be  89°  23'  50",  the  supplement  of 
which  90°  36'  10''  is  the  angle  ACB'.  Then,  in  the  triangle 
ABC, 

Sin  A  :  CB':  :Sin  C  :  AB'=86.45 

Ex.  2.  Given  the  angle  at  A  63°  35'  (Fig.  29.)  the  side 
b  64,  and  the  side  a  72  :  to  find  the  side  c.  and  the  angles  B' 
and  C 


TRIANGLES.  87 

«  :  Sin  A:  :6  :  Sin  B=52°  43'  25" 
Sin  A  :  a::SinC  :  c  =  72.05 

The  sum  of  the  angles  A  and  B  is  116°  20'  25',  the  sup- 
plement of  which  63°  39'  35"  is  the  angle  C. 

In  this  example  the  solution  is  7iot  ambi^uous^  hecause  the 
side  opposite  the  given  angle  is  longer  than  the  other  given 
side. 

Ex.  3.   In  a  triangle  of  which  the  angles  are  A,  B,  and  C, 
and  the  opposite  sides  a,  6,  and  c,   as  before  ;  if  the  angle  A 
be  121°  40',  the  opposite  side  a  68  rods,  and  the  side  6   47 
rods  ;  what  are  the  angles  B  and  C,  and  what  is  the  length  of 
the  side  c  '?  Ans.  B  is  36°  2'  4",  C  22°  17'  .56",  and  c  30.3. 


In  this  example  also,  the  solution  is  not  ambiguous,  because 
Case  III. 


the  gi-Dtn  angle  is  obtuse. 


\o3.  Given 
Two  sides,  and  ^  f    fi    1    ^  '^^^  remaining  side,  and 

The  included  angle,  )  X  The  other  two  angles. 

In  this  case,  the  angles  are  found  by  theorem  II.  (Art. 
144.)     The  required  side  may  be  found  by  theorem  I. 

In  making  the  solutions,  it  will  be  necessary  to  observe, 
that  by  subtracting  the  given  angle  from  180°,  the  sum  of  the 
other  two  angles  is  found  ;  (Art.  79.)  and,  that  adding  half 
the  difference  of  tzvo  quantities  to  their  half  sum  gives  the  great' 
er  quantity,  and  subtracting  the  half  difference  from  the  half 
sum  gives  the  less,  (Alg.  341.)  The  latter  proposition  may 
be  geometrically  demonstrated  thus  ; 

Let  AE  (Fig.  32.)  be  the  greater  of  two  magnitudes,  and 
BE  the  less.  Bisect  AB  in  D,  and  make  AC  equal  to  BE. 
Then 

AB  is  the  sum  of  the  tv/o  magnitudes  : 

CE     their  difference  ; 

DA  or  DB  half  their  su7n  ; 

IJE  or  DC  Juilf  their  difference  ; 
But  DA-}-DE=AE  the  greater  magnitude  ; 
And  DB  -  DE=BE  the  less, 

Ex.   1.  ]n  the  triangle  ABC  (Fig.  30.)  the  angle  A  is  giv- 


88  OBLIQUE  ANGLED 

en  26°  14'  the  side  b  39,  and  the  side  c  53  ;  to  find  the  an- 
gles B  and  C,  and  the  side  a. 

The  sum  of  the  sides  b  and  c  is         53-{-39=92, 
And  ihe'ir  differetice  53 — 39  =  14. 

The  sum  of  the  angles  B  and  C  =  180°-26°  14'  =  153°   46' 
And  half  the  sum  of  B  and  C  is  76°  53' 

Then,  by  theorem  If, 
(b+c)  :  (6-c)::Tani(B4-C)  :  Tan  K^^-C) 

'I  o  and  from  the  lialf  sum  76°  53' 

Adding  and  subtracting  the  half  difference  33     8    50 

We  have  the  greater  angle  HO     1    50 

And  the  less  angle  43  44  10 

As  the  greater  of  the  two  given  sides  is  c,  the  greater  an- 
gle is  C,  and  the  less  angle  B.  (Art.  149.) 

To  find  the  side  a,  by  theorem  1, 
Sin  B  :  ^>::Sin  A  :  a=24.94 

Ex.  2.  Given  the  angle  A  101°  30'  the  side  b  76,  and  the 
side  c  109  ;  to  find  the  angles  B  and  C,  and  the  side  «. 
B  is  30°  57f,  C  47°  32^,  and  a  144.8. 

Case  IV^ 

154.  Given  the  three  sides,  to  find  the  angles. 

In  this  case,  the  solutions  may  be  made,  by  drawing  a  per- 
pendicular to  the  longest  side,  from  the  opposite  angle.  This 
will  divide  the  given  triangle  into  two  rii^ht  angled  inanities. 
The  two  segments  may  be  found  by  theorem  111.  (Art.  145.) 
There  will  then  be  {Zjiven,  in  each  of  the  right  angled  trian- 
gles, the  hypothenuse  and  one  of  the  legs,  from  which  the 
angles  may  be  determined,  by  rectangular  trigonometry. 
(Art.  135.) 

Ex.  1.  Jn  the  triangle  ABC  (Fig.  31.)  the  side  AB  is  39, 
A.C  35,  and  BC  27.     What  are  the  angles  ? 

Let  a  perpendicular  be  drawn  from  C.  dividing  the  long- 


TRIANGLES.  89 

est  side  AB  into  the  two  segments  A?  and  BP.      Then  by 
theorem  III, 

AB  :  AC-|-BC::AC-BC  :  AP-BP 

As  the  longest  side  39  a.  c.  8.40894 

To  the  sum  of  the  two  others  62  1 .79239 

So  is  the  difference  of  the  latter  8  0.90309 


To  the  difference  of  the  segments         1 2.72        1 . 1 0442 


The  greater  of  the  two  segments  is  AP,  because  it  is  next 
the  side  AC,  which  is  greater  than  BC.   (Art.  146.) 

To  and  from  half  the  sum  of  the  segments  19.5 

Adding  and  subtracting  half  their  difference,  (Art.153.)  6.36 

We  have  the  greater  segment  AP  25.86 

And  the  less  BP  13.14 


7'hen,  in  each  of  the  right  angled  triangles  APC  and  BPC, 
we  have  given  the  hypothenuse  and  base  ;  and  by  art.  135, 

AC  :  R:  :AP  :  Cos  A=42°  21'  57' 
BC  :  R::BP:  Cos  B  =  60°  52'  42" 

And  subtracting  the  sum  of  the  angles  A  and  B  from  180°, 
we  have  the  remaining  angle  ACB  =  7G°  45'  21". 

Ex.    2.   If  the  three  sides  of  a  triangle  are  73,  96,  and  104  ; 
what  are  the  angles  ? 

Ans.  45°  41'  48  ",  61°  43'  27",  and  72°  34'  45". 

Examples  for  Practice* 

1.  Given  the  angle  A  54°  30',  the  angle  B  63°  10',  and  the 
side  a  164  rods ;  to  find  the  angle  C,  and  the  sides  b  and  c. 

2.  Given  the  angle  A  45°  6' the  opposite  side  a  93,  and  the 
side  b  108  ;  to  find  the  angles  B  and  C,  and  the  side  c. 

3.  Given  the  angle  A6T  24',  the  opposite  side  a  62,  and  the 
side  h  46  ]  to  find  the  angles  B  and  C,  and  the  side  c. 

4.  Given  the  angle  A  127°  42',  the  opposite  side  a  381,  and 
the  side  b  1 84 :  to  find  the  angles  B  and  C,  and  the  side  c. 

13 


90  OBLIQUE  ANGLED  TRIANGLES. 

5.  Given  the  side  b  58,  the  side  c  67,  and  the  included  an- 
gle A=36°  ;  to  tind  the  angles  B  and  C,  and  the  side  a, 

6.  Given  the  three  sides,631 ,2G8,and  546-,  to  find  the  angles. 

155.  The  three  theorems  demonstrated  in  this  section, 
have  been  here  applied  to  oblique  ansrled  triangles  only.  But 
they  are  equally  applicable  to  risrhl  a/i^/erZ  triangles. 

Thus,  in  the  triangle  ABC,  (Fig.  17.)  according  to  theo- 
rem I,  (Art.  143.) 

SinB:  AC::Sin  A:BC 

This  is  the  same  proportion  as  one  stated   in  art.  134,  ex- 

ceptthat,  in  the  first  term  here,  the  sine  of  Bis  substituted  for 

radius.     But,  as  B  is  a  right  angle,  its  sine  is  equal  to  radius, 

(Art.  95.) 

Again,  in  the  triangle  ABC,  (Fig.  21.)  by  the  same  theorem ; 

Sin  A  :BC::SinC  :  AB 

This  is  also  one  of  the  proportions  in  rectangular  trigo- 
nometry, when  the  hypothenuse  is  made  radius. 

The  other  two  theorems  might  be  applied  to  the  solution 
of  right  angled  triangles.  But,  when  one  of  the  angles  is 
known  to  be  a  right  angle,  the  methods  explained  in  the  pre- 
ceding section,  are  much  more  simple  in  practice.* 

*  For  the  application  of  Trigonometry  to  the  Mensuration  of  Heights  and 
Distances,  see  Navigation  and  Surveying. 


SECTION  V. 


GEOMETRICAL  CONSTRUCTION  OF  TRIANGLES, 
BY  THE  PLANE  SCALE. 


.  .  P  ^^O  facilitate  the  construction  of  geometrical 
-*-  figures,  a  nunnber  of  graduated  lines  are  put 
upon  the  common  two  feet  scale  ;  one  side  of  which  is  call- 
ed the  Plane  Scale,  and  the  other  side,  Gunter^s  Scale.  The 
most  important  of  these  are  the  scabies  of  equal  parts,  and  the 
line  o( chords.  In  forming  a  given  triangle,  or  any  other  right 
lined  figure,  the  parts  which  must  be  made  to  agree  with  the 
conditions  proposed,  are  the  lines,  and  the  angles.  For  the 
former,  a  scale  of  equal  parts  is  used  ;  for  the  latter,  a  line  of 
chords. 

157.  The  line  on  the  upper  side  of  the  plane  scale,  is  di- 
vided into  inches  and  tenths  of  an  inch.  Beneath  this,  on  the 
left  hand,  are  two  diagonal  scai\es  of  equal  parts,*  divided  into 
inches  and  half  inches,  by  perpendicular  lines.  On  the  lar- 
ger scale,  one  of  the  inches  is  divided  into  tenths,  by  lines 
which  pass  ohliqueli^  across,  so  as  to  intersect  the  parallel  lines 
which  run  from  right  to  left.  The  use  of  the  oblique  lines  is 
to  measure  hundredths  of  an  inch,  by  inclining  more  and 
more  to  the  right,  as  they  cross  each  of  the  parallels. 

To  take  off.  for  instance,  an  extent  of  3  inches,  4  tenths, 
and  6  hundredths ; 

Place  one  foot  of  the  compasses  at  the  intersection  of  the 
perpendicular  line  marked  3  with  the  parallel  line  marked  6, 
and  the  other  foot  at  the  intersection  of  the  latter  with  the 
oblique  line  marked  4, 

The  other  diagonal  scale  is  of  the  same  nature.  The  di- 
visions are  smaller,  and  are  numbered  from  left  to  right. 

158.  In  geometrical  constructions,  what  is  often  required, 
is  to  make  a  figure,  not  equal  to  a  given  one,  but  only  similar* 
Now  figures  are  similar  which   have  equal   angles,   and   the 

*  These  lines  are  not  represeated  in  the  plate,  as  the  learner  is  supposed  to 
have  the  scale  before  him. 


92  GEOxMETRICAL  CONSTRUCTION    * 

sides  about  the  equal  angles  proportioned,  (Euc.  Dei.  1.  6.; 
Thus  a  hind  surveyor,  in  plotting  a  field,  makes  the  several 
lines  in  his  plan  to  have  the  same  proportion  to  each  other, 
as  the  sides  ol'  the  field.  For  this  purpose,  a  scale  of  equal 
parts  may  he  used,  of  any  dimensions  whatever.  If  the  sides 
ofthc  field  are  2,  5,  7,  and  10  rods,  and  the  lines  in  the  plan 
are  2,  5,  7,  and  10  inches^  and  if  the  angles  are  the  same  in 
each,  tlie  figures  are  similar.  One  is  a  copy  of  the  other, 
upon  a  smaller  scale. 

So  any  two  right  lined  figures  are  similar,  if  the  angles  are 
the  same  in  both,  and  if  the  number  of  smaller  parts  in  each 
side  of  one,  is  equal  to  the  number  of  larger  parts  in  the  cor- 
responding sides  of  the  other.  The  several  divisions  on  the 
scale  of  equal  parts  may,  therefore,  be  considered  as  repre- 
senting any  measures  of  length,  as  feet,  rods,  miles,  &;c.  All 
that  is  necessary  is,  that  the  scale  be  not  changed,  in  the  con- 
struction of  the  same  figure ;  and  that  the  several  divisions 
and  subdivisions  be  proptrly  proportioned  to  each  other.  If 
the  larger  divisions,  on  the  diagonal  scale,  are  units,  the  smal- 
ler ones  are  tenths  and  hundredths.  If  the  larger  are  tens, 
the  smaller  are  units  and  tenths. 

159.  In  laying  down  an  angle,  of  a  given  number  of  de- 
grees, it  is  necessary  to  measure  it.  Now  the  proper  measure 
of  an  angle  is  an  arc  of  a  circle.  (Art.  74.)  And  the  measure 
of  an  arc,  where  the  radius  is  given,  is  hs  chord.  For  the 
chord  is  the  distance,  In  a  straight  line,  from  one  end  of  the 
arc  to  the  other.  Thus  the  chord  AB  (Fig.  33.)  is  a  meas- 
ure of  the  arc  .ADB,  and  of  the  angle  ACB. 

To  form  the  line  of  chords,  a  circle  is  described,  and  the 
lengths  of  its  chords  determined  for  every  degree  of  the  quad- 
rant. These  measures  are  put  on  the  plane  scale,  on  the  line 
marked  CHO. 

160.  The  chord  of  60°  is  equal  to  radius.  (Art,  95.)  In 
hiying  down  or  measuring  an  angle,  therefore,  an  arc  must 
be  draun,  with  a  radius  which  is  equal  to  the  extent  from  0 
to  60  on  the  line  of  chords.  There  are  generally  on  the  scale, 
two  lines  of  chords.  Either  of  these  may  be  used  ;  but  the 
angle  must  be  measured  by  the  same  line  from»which  the 
radius  is  taken. 

161.  To  make  an  angle,  then,  of  a  given  number  of  de- 
grees ;  From  one  end  of  a  straight  line  as  a  centre,  and  with 
a  radius  equal  to  the  chord  of  60°  on  the  line  of  chords,  de- 
scribe an  arc  of  a  circle  cuttina;  the  straight  line.     From  the 


OF  TRIANGLES. 

poiot  of  intersection,  extend  the  chord  of  the  given  number 
of  degrees,  applying  the  other  extremity  to  the  arc ;  and 
through  the  place  of  meeting,  draw  the  other  line  from  the 
angular  point. 

Jf  the  given  angle  is  obtuse,  take  from  the  scale  the  chord 
of  Art//* the  number  of  degrees,  and  apply  it  twice  to  the  arc. 
Or  make  use  of  the  chords  of  any  two  arcs  whose  sum  is  equal 
to  the  given  number  of  degrees. 

A  right  angle  may  be  constructed,  by  drawing  a  perpen- 
dicular without  using  the  line  of  chords. 

Ex.  I.  To  make  an  angle  of  32  degrees.  (Fig.  33.)  With 
the  point  C,  in  the  line  CH,  for  a  centre,  and  with  the  chord 
of  60°  for  radius,  describe  the  arc  ADF.  Extend  the  chord 
of32''  from  A  to  B  ;  and  through  B,  draw  the  line  BC. 
Then  is  ACB  an  an^le  of  32  degrees. 

2.  To  make  an  angle  of  140  degrees.  (Fig.  34.)  On  the 
line  CH,  with  the  chord  of  60°,  describe  the  arc  ADF ;  and 
extend  the  chord  of  70°  from  A  to  D,  and  from  D  to  B.  The 
arc  ADB=70°x2  =  140.° 

On  the  other  hand, 

162.  To  measure  an  angle  ;  On  the  angular  point  as  a  cen- 
tre, and  with  the  chord  of  60°  for  radius,  describe  an  arc  to 
cut  the  two  lines  which  include  the  angle.  The  distance  be- 
tween the  points  of  intersection,  applied  to  the  line  of  chords, 
will  give  the  measure  of  the  angle  in  degrees.  If  the  angle 
be  obtuse,  divide  the  arc  into  two  parts. 

Ex.  1.  To  measure  the  angle  ACB.  (Fig:  33.)  Describe 
the  arc  ADF  cutting  the  lines  CH  and  CB.  The  distance 
AB  will  extend  32°  on  the  line  of  chords. 

2.  To  measure  the  angle  ACB.  (Fig.  34.)  Divide  the 
arc  ADB  into  two  parts,  either  equal  or  unequal,  and  meas- 
ure each  part,  by  applying  its  chord  to  the  scale.  The  sum 
of  the  two  will  be  140°. 

163.  Besides  the  lines  of  chords,and  of  equal  parts,  on  the 
plane  scale  ;  there  are  also  lines  of  natural  sines,  tangents,  and 
secants,  marked  Sin.  Tan.  and  Sec.  oi  semitangents,  marked 
S.  T.  o(  longitude,  marked  Lon.  or  M.  L.  of  rhumbs,  marked 
Rhu.  or  Rum.  he.  These  are  not  necessary  in  trigonomet- 
rical constructions.  Some  of  them  are  used  in  Navigation  ; 
and  some  of  them,  in  the  projections  of  the  Sphere. 

164.  In  Navigation,  the  quadrant,  instead  of  being  gradua- 
ted in  the  usual  manner,  is  divided  into  eight  portions,  called 


94  GEOMETRICAL  CONSTRUCTION 

Rhumbs,     The  Rhumb  line^  on  the  scale,  is  a  line  ol  chords, 
divided  into  rhumbs  and  quarter-rhumbs,  instead  of  degrees. 

165.  The  line  o{  Longitude  is  intended  to  show  the  num-= 
ber  ot" geographical  miles  in  a  degree  of  longitude,  at  different 
distances  from  the  equator.  It  is  placed  over  the  line  of 
chords,  with  the  numbers  in  an  inverted  order  :  so  that  the  fig- 
ure  above  shows  the  length  of  a  degree  of  longitude,  in  any 
latitude  denoted  by  the  figure  below.*  Thus  at  the  equator, 
where  the  latitude  isO,  a  degree  of  longitude  is  GO  geographi- 
cal miles.  In  latitude  40,  it  is  46  miles;  in  latitude  60,  30 
miles,  &ic. 

166.  The  graduation  on  the  line  of  secants  begins  where 
the  line  of  sines  ends  For  the  greatest  sine  is  only  equal  to 
radius;  but  the  secant  of  the  least  arc  is  greater  than  radius. 

167.  The  semitangents  are  the  tangents  of  half  the  given 
arcs.  Thus  the  semitangent  of  20°  is  the  tangent  of  10°. 
The  line  of  semitangents  is  used  in  one  of  the  projections  of 
the  sphere. 


168.  In  the  construction  of  triangles^  the  sides  and  angles 
which  are  given,  are  laid  down  according  to  the  directions  in 
arts.  158,  161.  The  pans  r eg iiired  ure  then  measured,  ac- 
cording to  arts.  158,  162.  The  following  problems  corres- 
pond with  the  four  cases  of  oblique  angled  triangles;  (Art. 
148.)  but  are  equally  adapted  to  right  angled  triangles. 

169.  Pkob.  1.  The  angles  and  one  side  of  a  triangle  being 
given  ;  to  find,  by  construction,  the  other  two  sides. 

Draw  the  given  side.  From  the  ends  of  it,  lay  off  two  of 
the  given  angles.  Extend  the  other  sides  till  they  intersect  j 
and  then  measure  their  lengths  on  a  scale  of  equal  parts. 

Ex.  1.  Given  the  side  6  32  rods,  (Fig.  27.)  the  angle  A 
56°  20',  and  the  angle  C  49°  10  ;  to  construct  the  triangle, 
and  find  the  length.:3  of  the  sides  a  and  r. 

Their  lengths  will  be  25  and  27i. 

2.  In  a  right  angled  triangle,  (Fig.  17.)  given  the  hypoth- 
enuse  90,  and  the  angle  A  32°  20',  to  find  the  base  and  per- 
pendicular. 

The  length  of  AB  will  be  76,  and  of  BC  48. 

*  Sometimes  the  line  of  loDgitude  is  placed  tt/Lder  the  line  of  chord?. 


OF    TRIANGLES.  95 

3.  Given  the  side  AC  68,  the  angle  A  124°,  and  the  angle 
C  37°  :  to  construct  the  triangle. 

170  Prob.  II.  Two  sides  and  an  opposite  angle  being  giv- 
en, to  find  the  remaining  side,  and  the  other  two  angles. 

Draw  one  of  the  given  sides ;  from  one  end  of  it,  lay  off 
the  given  angle  ;  and  extend  a  line  indefinitely  for  the  required 
side.  From  the  other  end  of  the  first  side,  with  the  remain- 
ing given  side  for  radius,  describe  an  arc  cutting  the  indefinite 
line.  The  point  of  intersection  will  be  the  end  of  the  requir- 
ed side. 

If  the  side  opposite  the  given  angle  be  less  than  the  other 
given  S'de,  the  case  will  be  ambiguous    (Art.  152.) 

Ex.  1.  Given  the  angle  A  63°  35'  (Fig.  29.)  the  side  b  32, 
and  the  side  a  36. 

The  side  AB  will  be  36  i^.early,  the  angle  B  52°  45V  and 
C  63°  391' 

2.  Given  the  angle  A  (Fig.  26.)  35°  20';  the  opposite  side 
a  25.  and  the  side  h  35. 

Draw  the  side  b  35,  make  the  angle  A  35°  20',  and  extend 
AH  indefinitely.  From  C  wi».h  radius  25,  describe  an  arc 
cutting  AH  in  B  and  B'.  Draw  CB  and  CB',  and  two  trian- 
gles will  be  formed,  ABC  and  AB'C,  each  corresponding 
with  the  conditions  of  the  problem. 

3.  Given  the  angle  A  1 16°,  the  opposite  side  a  38,  and  the 
side  b  26  ;  to  construct  the  triangle. 

171.  Prob.  Hi.  Two  sides  and  the  included  angle  being 
given  ;  to  find  the  other  side  and  angles. 

Draw  one  of  the  given  sides.  From  one  end  of  it  lay  off 
the  given  angle,  and  draw  the  other  given  side.  Then  con- 
nect the  extremities  of  this  and  the  first  line. 

Ex.  i:  Given  the  angle  A  (Fig.  30.)  26°  14',  the  side  // 
78,  and  the  side  c  106;  to  find  B,  C.  and  a. 

The  side  a  will  be  50.  the  angle  B  43°  44',  and  C  110°  2. 

2.   Given  A  86°,  b  65,  and  c  83 ;  to  find  B,  C,  and  a. 

172.  Prob.  IV.  The  three  sides  being  given;  to  find  the 
angles. 

Draw  one  of  the  sides,  and  from  one  end  of  it,  with  an  ex- 
tent equal  to  the  second  side,  describe  an  arc.  From  the 
other  end,  with  an  extent  equal  to  the  third  side,  describe  a 
second  arc  cutting  the  first;  and  from  the  point  of  intersec- 
tion draw  the  two  sides.  (Euc.  22.  1.) 

Ex.  1.  Given  AB  (Fig.  31.)  78,  AC  70,  and  BC,  54 ;  to 
find  the  angles. 


3C     GEOMETRICAL  CONSTRUCTION  OF  TRIANGLES. 

The  angles  will  be  A  42°  22',  B  60^^  52|',  and  C  76°  451'. 

2.  Given,  the  three  sides  68,  39,  and  40 ;  to  find  the  an- 
gles. 

173.  Any  right  lined  figure  whatever,  whose  sides  and  an- 
gles are  given,  may  be  constructed,  by  laying  down  the  sides 
from  a  scale  of  equal  parts,  and  the  angles  from  a  line  of 
chords. 

Ex.  Given  the  sides  AB  (Fig.  35.)  =20,  BC=22,  CD= 
30,  DE  =  12;  and  the  angles  B  =  102°,  C  =  130%  D  =  108°, 
to  construct  the  figure. 

Draw  the  side  AB  =  20,  make  the  angle  B  — 102°,  draw 
BC=22,  make  C  =  130»,  draw  CD=30,  make  D=108% 
draw  DE  =  12,  and  connect  E  and  A. 

The  last  line  EA  may  be  measured  on  the  scale  of  equal 
parts;  and  the  angles  E  and  A,  by  a  line  of  chords. 


SECTION  VI. 
DESCRIPTION  AND  USE  OF  GUNTER'S  SCALE. 


\  I  "4  A^  expeditious  method  of  solving  the  prob- 
^  RT.  .  -^!^|gjj^g  in  trigonometry,  and  making  other  loga- 
rithmic calculations,  in  a  mechanical  way,  has  been  contrived 
by  Mr.  Edmund  Gunter.  The  logarithms  of  numbers,  of 
sines,  tangents,  &:c.  are  represented  by  lines.  By  means  of 
these,  multiplication,  division,  the  rule  of  three,  involution, 
evolution,  &;c.  may  be  performed  much  more  rapidly,  than  in 
the  usual  method  by  figures. 

The  logarithmic  lines  are  generally  placed  on  one  side  only 
of  the  scale  in  common  use.     They  are, 

A  line  of  artificial  Sines  divided  into  Rhumbs,  and  mark- 
ed S.  R. 
A  line  of  artificial  Tangents,  do.  T,  R. 
A  line  of  the  logarithms  o(  numbers,  Num. 
A  line  of  artificial  Sines,  to  every  degree.  SIN. 
A  line  of  artificial  Tangents,  do.  TAN. 
A  line  of  Versed  Sines,  V.  S- 

To  these  are  added  a  hue  of  equal  parts,  and  a  line  o(  Me- 
ridional Parts,  which  are  not  logarithmic.  The  latter  is  used 
in  Navigation. 

The  Line  of  JVumbers. 

175.  Portions  of  the  line  oi  Numbers,  are  uitended  to  rep- 
resent the  logarithms  of  the  natural  series  of  numbers  2,  3,  4, 
5,  &c. 

The  logarithms  of  10,  100,  1000,  &c.  are  1,  2,  3,  &c. 
(Art.  3.) 

Tf  then,  the  log.  of  10  be  represented  by  a  line  of  1  foot; 
the  log.  of  100  will  be  repres'd  by  one  of  2  feet ; 
the  log.  of  1000  by  one  of  S  feet; 

the  lengths  of  the  several  lines  being  proportional  to  the  cor- 
responding logarithms  in  the  tables.  Portions  of  a  foot  will 
represent  the  logarithms  of  numbers   between   1  and  10; 

14 


98  rillGONOMETRV- 

and  poitions  of  a  line  2  feet  long,  the  logarithmb  of  numbers 
between  1  and  100. 

On  Gunter's  scale,  the  line  of  the  logarithms  of  numbers 
begins  at  a  brass  pin  on  the  left,  and  the  divisions  are  num- 
bered 1,  2,  3,  (foe.  to  another  pin  near  the  middle.  From  this 
the  numbers  are  repeated,  2,  3,  4,  &;c.  which  may  be  read 
^0,  30,  40,  &IC.  The  logarithms  of  numbers  between  1  and 
10  are  represented  by  portions  of  the  first  half  of  the  line  ; 
and  the  logarithms  of  numbers  between  10  and  100,  by  por- 
tions greater  than  half  the  line,  and  less  than  the  whole. 

17C.  The  logarithm  of  1,  which  is  0,  is  denoted,  not  by 
any  extent  of  line,  but  by  a  point  under  1,  at  the  commence- 
ment of  the  scale.  The  distances  from  this  point  to  differ- 
ent parts  of  the  line,  represent  other  logarithms,  of  which 
the,  figures  placed  over  the  several  divisions  are  the  natural 
/lumbers.  For  the  intervening  logarithms,  the  intervals  be- 
tween the  ligures,  are  divided  into  tenths,  and  sometimes  in- 
to smaller  portions.  On  the  right  hand  half  of  the  scale,  as 
the  divisions  which  are  numbered  are  tens,  the  subdivisions 
are  units. 

Ex.  1.  To  take  from  the  scale  the  logarithm  of  3.6  ;  set 
one  foot  of  the  compasses  under  1  at  the  beginning  of  the 
scale,  and  extend  the  other  to  the  6th  division  after  the  first 
figure  3. 

2.  For  the  logarithm  of  47  ;  extend  from  1  at  the  begin- 
ning, to  the  7th  subdivision  after  the  second  figure  4.* 

177.  It  will  be  observed,  that  the  divisions  and  subdivi- 
sions decrease,  from  left  to  right;  as  in  the  tables  of  loga- 
rithms, the  differences  decrease.  The  difference  between 
the  logarithms  of  10  and  100  is  no  greater,  than  the  differ- 
ence between  the  logarithms  of  1  and  10. 

178.  The  line  of  numbers,  as  it  has  been  here  explained, 
fijrnishes  the  logarithms  of  all  numbers  between  1  and  100. 

And  if  the  indices  of  the  logarithms  be  neglected,  the  same 
scale  may  answer  for  all  numbers  whatever.  For  the  deci- 
mal part  of  the  logarithm  of  any  number  is  the  same,  as  that 
of  the  number  multiplied  or  divided  by  10,  100,  8zc»  (Art. 
14.)  In  logarithmic  calculations,  the  use  of  the  indices  is  to 
determine  the  distance  of  the  several  figures  of  the  natural 
numbers  from  the  place  of  units.  (Art.  11.)  But  in  those 
cases  in  which  the  logarithmic  line  is  commonly  used,  it  will 

*  If  the  compasses  will  not  reach  the  distance  required  ;  first  open  them  so 
as  to  take  afffijalff  or  any  part  of  the  distance,  and  then  the  remaining  part. 


GUNTER-S  SCALE.  d9 

not  generally  be  difficult  to  determine  the  local  value  of  the 
figures  in  the  result. 

179.  We  may,  therefore,  consider  the  point  under  1  at  the 
left  hand,  as  representing  the  logarithm  of  1,  or  10,  or  100  ; 
or  y*;j,  or  y^oi  <^c.  for  the  decimal  part  of  the  logarithm  of 
each  of  these  is  0.  But  if  the  first  1  is  reckoned  10,  all  the 
succeeding  numbers  must  also  be  increased  in  a  tenfold  ra- 
tio ;  so  as  to  read,  on  the  first  half  of  the  line,  20,  30,  40,  &.c. 
and  on  the  other  naif,  200,  300,  kc. 

The  whole  extent  of  the  logarithmic  line, 
is  from  1       to  ICO,  or  from  0.1       to  10. 

or  from  10    to  1000,  or  from  0.01     to  1, 

or  from  100  to  10000,  kc.       or  from  O.OOI  to  0.1,  &c. 

Different  values  may,  on  diflerent  occasions,  be  assigned  to 
the  several  numbers  and  subdivisions  marked  on  this  line. 
But  for  any  one  calculation,  the  value  must  remain  the  same. 

Ex.  Take  from  the  scale  3C5. 

As  this  number  is  between  10  and  1000,  let  the  1  at  the 
beginning  of  the  scale,  be  reckoned  10.  Then,  from  this 
point  to  the  second  3  is  300  ;  to  the  6th  dividing  stroke  is 
60  ;  and  half  way  trom  this  to  the  next  stroke  is  o. 

ISO.  Multiplication,  division,  (Sic.  are  performed  by  the 
line  of  numbers,  on  the  same  principle,  as  by  common  loga- 
rithms.    Thus, 

To  multiplt/  by  this  line,  add  the  logarithms  of  the  two 
factors;  (Art.  37.)  that  is,  takeoff,  with  the  compasses,  that 
length  of  line  which  represents  the  logarithm  of  one  of  the 
factors,  and  apply  this  so  as  to  extend  forward  from  the  end 
of  that  which  represents  the  logarithm  of  the  other  factor. 
The  sum  of  the  two  will  reach  to  the  end  of  the  line  reprc- 
senting  the  logarithm  of  the  product. 

Ex.  Multiply  9  into  8.  The  extent  from  1  to  8,  added  to 
that  from  1  to  0,  will  be  equal  to  the  extent  from  1  to  72  the 
product. 

181.  To  divide  by  the  logarithmic  line,  subtract  the  loga- 
rithm of  the  divisor  from  that  of  the  dividend;  (Art.  41.)  that 
is,  take  off  the  logarithm  of  the  divisor,  and  this  extent  set 
back  from  the  end  of  the  logarithm  of  the  dividend,  will  reach 
to  the  logarithm  of  the  quotient. 

Ex.  Divide  42  by  7.  The  extent  from  1  to  7,  set  back 
from  42,  will  reach  to  6  the  quotient. 

182.  Involution  is  performed  in  logarithms,  by  multiplying 
<he  logarithm  of  the  quantity  into  the  index  of  (he  power: 


IQO  TRIGONOMETKY. 

(Art.  45.)  that  is,  by  repeating  the  logarithms  as  many  times 
as  there  are  units  in  the  index.  To  involve  a  quantity  on  the 
scale,  then,  take  in  the  compasses  the  linear  logarithm,  and 
double  it,  treble  it,  <S:c.  according  to  the  index  of  the  proposed 
power. 

Ex.  1.  Required  the  square  of  9.  Extend  the  compas- 
ses from  1  to  9.  Twice  this  extent  will  reach  to  81  the 
square. 

2.  Required  the  cube  of  4.  The  extent  from  1  to  4,  re- 
peated three  times,  will  reach  to  64  the  cube  of  4. 

183.  On  the  other  hand,  to  perform  evolution  on  the  scale  ; 
take  half,  one  third,  &;c.  of  the  logarithm  of  the  quantity,  ac- 
cording to  the  index  of  the  proposed  root. 

Ex.  1.  Required  the  scjuare  root  of  49.  Half  the  extent 
from  1  to  49,  will  reach  from  1  to  7  the  root. 

2.  Required  the  cube  root  of  27.  One  third  the  distance 
from  1  to  27,  will  extend  from  I  to  3  the  root. 

184.  The  Rule  of  Three  may  be  performed  on  the  scale, 
in  the  same  manner  as  in  logarithms,  by  adding  the  two  mid- 
dle terms,  and  from  the  sum,  subtracting  the  first  term,  (v/^rt. 
52.)  But  it  is  more  convenient  in  practice  to  begin  by  sub- 
tracting the  first  term  from  one  of  the  others.  If  four  quan- 
tities are  proportional,  the  quotient  of  the  first  divided  by  the 
second,  is  equal  to  the  quotient  of  the  third  divided  by  the 
fourth.  (Alg.  364.) 

a     <  II     b 

Thus  if  a  :  bWc  t  </,  then  t=-%,  and -=y  (Alg.  380.) 

But  in  logarithms,  subtraction  takes  the  place  of  division; 
so  that, 

log.  a  — log.6  =  log.  c— log.a.  Orlog.  <z -log.  c=log.6  — log.c?. 
Hence, 

185.  On  the  scale,  the  difference  beizveen  the  frst  and  se- 
cond terms  of  a  proportion,  is  equal  to  the  difference  between 
the  third  and  fourth.  Or,  the  diflference  between  the  first  and 
third  terms,  is  equal  to  the  difference  between  the  second 
and  fourth. 

The  difference  between  the  two  terms  is  taken,  by  extend- 
ing the  compasses  from  one  to  the  other.  If  the  second 
term  be  greater  than  the  first;  the  fourth  must  be  greater 
than  the  third ;  if  less,  less.  (Alg.  395.*)  Therefore  if  the 
compasses  extend  forzvard  from  I  ft  to  right,  that  is,  from  a 

*  Eijc.  14.  .'■). 


GUNTER'S  SCALE.  lOi 

less  number  to  a  greater,  from  the  first  term  to  the  second ; 
they  must  also  extend  forward  from  the  third  to  the  fourth. 
But  if  they  extend  backward^  from  the  tirst  term  to  the  sec- 
ond ;  they  must  extend  the  same  way,  from  the  third  to  the 
fourth. 

Ex.  I.  In  the  proportion  3  :  8:  :12  :  32,  the  extent  from 
3  to  8,  will  reach  from  12  to  32;  Or,  the  extent  from  3  to 
12,  will  reach  from  8  to  32. 

2.  If  54  yards  of  cloth  cost  48  dollars,  what  will  18  yards 

cost?  54  :  48::i8  :  i6 

The  extent  from  54  to  48,  will  reach  backwards  from  18 
to  i6. 

3.  If  63  gallons  of  wine  cost  81  dollars,  what  will  35  gal- 
lons cost?  63  :  01  ::35  :  45 

The  extent  from  63  to  81,  will  reach  from  35  to  45. 


The  Line  of  Sines. 

186.  The  line  on  Gunter's  scale  marked  SIN.  is  a  line  of 
logarithmic  sines,  made  to  correspond  with  the  line  of  num- 
bers. The  whole  extent  of  the  line  of  numbers,  (Art.  179.) 
is  from  1  to  100,  whose  logs,  are  0.00000  and  2.00000, 
or  from  10  to  1000,  whose  logs,  are  1.00000  and  3.00000, 
or  from  100  to  10000,  whose  logs,  are  2.00000  4.00000, 
the  difference  of  the  indices  of  the  two  extreme  logarithms 
being  in  each  case  2. 

Now  the  logarithmic  sine  of  0^  34'  22'' 41'"  is  8.00000 
And  the  sine  of  90°  (Art.  95.)  is  10.00000 

Here  also  the  difference  of  the  indices  is  2.  If  then  (he 
point  directly  beneath  one  extremity  of  the  line  of  numbers, 
be  marked  for  the  sine  of  0°  34' 22'' 41'";  and  the  point  be- 
neath the  other  extremity,  for  the  sine  of  90^  ;  the  interval 
may  furnish  the  intermediate  sines  ;  the  divisions  on  it  being 
made  to  correspond  with  the  decimal  part  of  the  logarithmic 
sines  in  the  tables.* 


'  To  represent  the  sines  less  than  34'  ^2"  41'  ,  the  scale  must  be  extended 
on  the  left  indefinitely.  For,  as  the  sine  of  an  arc  approaches  to  0,  its  1o<j;h- 
rithm,  whi'^h  i«:  negative,  inoreasos  withont  liniit-  ( \n.  1.').', 


iOii  TRIGONOMETRY. 

The  first  dividing  stroke  in  the  hne  of  Sines  is  generally  at 
0°  40',  a  httle  farther  to  the  right  than  the  beginning  of  the 
line  of  Numbers.  The  next  division  is  at  0°  50' ;  then  be- 
gins the  numbering  of  the  degrees,  1,2,  3,  4,  &c.  from  left  to 
right. 

The  line  of  Tangents, 

187.  The  first  45  degrees  on  this  line  are  numbered  from 
left  to  right,  nearly  in  the  same  manner  as  on  the  line  of 
Sines. 

The  logarithmic  tangent  of  0°  34'  22''  35'"  is    8.00000 
And  the  tangent  of  45°,  (Art.  95.)  is  10.00000 

The  difference  of  the  indices  being  2,  45  degrees  will 
reach  to  the  end  of  the  line.  For  those  above  45°  the  scale 
ought  to  be  continued  much  farther  to  the  right.  Bu^;  as  this 
would  be  inconvenient,  the  numbering  of  the  degrees,  after 
reaching  45,  is  carried  hack  from  right  to  left.  The  same 
dividing  stroke  answers  for  an  arc  and  its  complement^  one 
above  and  the  other  below  45°.     For,  (Art.  93.  Proper.  9.^ 

tan  :  R::R  :  cot 

In  logarithms,  therefore,  (Art.  184. 
tan-R=Pt— cot. 

That  is,  the  difference  between  the  tangent  and  radius,  is 
equal  to  the  difference  between  radius  and  the  cotangent :  in 
other  words,  one  is  as  much  greater  than  the  tangent  of  45°, 
as  the  other  is  less.  In  taking,  then,  the  tangent  of  an  arc 
greater  than  45°,  we  are  to  suppose  the  distance  between  45 
and  the  division  markea  with  a  given  number  of  degrees,  to 
be  added  to  the  whole  line,  in  the  same  manner  as  if  the  line 
were  continued  out.  In  working  proportions,  extending  the 
compasses  hack^  from  a  less  number  to  a  greater,  must  be 
considered  the  same  as  carrying  them  /c)?*ri'«rc/  in  other  cases. 
See  art.  185. 

Trigonometrical  Proportions  on  the  Scale* 

188.  In  working  proportions  in  trigonometry  by  the  scale  ; 

the   extent   from  the    first  term  to  the  middle  term  of  the  same 


GUNTER'S  SCALE.  103 

uumej  will  reach  from  the  other  middle  term  to  the  fourth  term* 
(Art.  185.) 

In  a  trigonometrical  proportion,  two  of  the  terms  are  the 
lengths  of  sides  of  the  given  triangle  ;  and  the  other  two  are 
tahular  sines,  tangents,  &c.  The  former  are  to  be  taken 
from  the  line  of  numbers  ;  the  latter,  from  the  lines  of  loga- 
rithmic sines  and  tangents.  If  one  of  the  terms  is  a  secant, 
the  calculation  cannot  be  made  on  the  scale,  which  has  com- 
monly no  line  of  secants-  It  must  be  kept  in  mind  that  ra- 
dius is  equal  to  the  sine  of  90°,  or  to  the  tangent  of  45°. 
(Art.  95.)  Therefore,  whenever  radius  is  a  term  in  the  pro- 
portion, one  foot  of  the  compasses  must  be  set  on  the  end  of 
the  line  of  sines  or  of  tangents. 

189.  The  following  examples  are  taken  from  the  propor- 
tions which  have  already  been  solved  by  numerical  calcu- 
lation. 

Ex.  1.  In  Case  I  of  right  angled  triangles,  (Art.  134.  ex.  1.) 

R  :  45;:  Sin  32*20'  :  24 

Here  the  third  term  is  a  sine  ;  the  first  term  radius  is,  there- 
fore, to  be  considered  as  the  sine  of  90^  Then  the  extent 
from  90°  to  32'='  20'  on  the  line  of  sines,  will  reach  from  45 
to  24  on  the  line  of  numbers.  As  the  compasses  are  set  bade 
from  90°  to  32°  20' ;  they  must  also  be  set  back  from  45. 
(Art.  185.) 

2.  In  the  same  case,  if  the  base  be  made  radius,  (Page  60.) 

R  :  38::  Tan  32°  20'  :  24 

Here,  as  the  third  term  is  a  tangent,  the  first  term  radius  is 
to  be  considered  the  tangent  of  45°.  Then  the  extent  from 
45°  to  32^  20'  on  the  line  of  tangents,  will  reach  from  38  to 
24  on  the  line  of  numbers. 

3.  If  the  perpendicular  be  made  radius,  (Page  60.) 

R  :  24:  :Tan  57^40'  :  38 

Ihe  extent  from  45°  to  57°  40'  on  the  line  of  tangents,  will 
reach  from  24  to  38  on  the  line  of  numbers.  For  thetangent 
of  57°  40'  on  the  scale  look  for  its  complement  32°  20'.  (Art. 
187.)     In  this  example,  although  the  compasses  extend  hack 


104  TKlGONOiMETKY. 

from  45°  to  57°  40' ;  yet,  as  this  is  from  a  less  number  to  a 
greater^  they  must  extend  forward  on  the  line  of  numbers. 
(Art.  185.  187.) 

4.  In  art.  135,  35  :  R:  :26  :  Sin  48° 
The  extent  from  35  to  26  will  reach  from  90°  to  48°. 

5.  In  art.  136,  R  :  48:  iTan  27°i  :  24^ 
The  extent  from  45°  to  27°},  will  reach  from  48  to  24J. 

6.  Inart.l50,ex.  1.       Sin74°30'  :  32:  :Sin66°20'  :  27|. 

For  other  examples,  see  the  several  cases  in  Sections  III. 
and  IV. 

190.  Though  the  solutions  in  trigonometry  may  be  effect- 
ed by  the  logarithmic  scale,  or  by  geometrical  construction, 
as  well  as  by  arithmetical  computation  ;  yet  the  latter  method 
is  by  far  the  most  accurate.  The  first  is  valuable  principally 
for  the  expedition  with  which  the  calculations  are  made  by  it. 
The  second  is  of  use,  in  presenting  the  form  of  the  triangle 
to  the  eye.  But  the  accuracy  which  attends  arithmetical  op- 
erations, is  not  to  be  expected,  in  taking  lines  from  a  scale 
with  a«pair  of  compasses.* 

*  See  INotr  H, 


SECTION  VII.* 


THE  FIRST  PRINCIPLES  OF  TRIGONOMETRI 
CAL  ANALYSIS. 


A  IQI  T^  the  preceding  sections,  sines,  tangents,  and 
secants,  have  been  employed  in  calculating 
the  sides  and  angles  of  triangles.  But  the  use  of  these  lines 
is  not  confined  to  this  object.  Important  assistance  is  derived 
from  them,  in  conducting  many  of  the  investigations  in  the 
higher  branches  of  analysis,  particularly  in  physical  astrono- 
my. It  does  not  belong  to  an  elementary  treatise  of  trigonom- 
etry, to  prosecute  these  inquiries  to  any  considerable  extent. 
But  this  is  the  proper  place  for  preparing  theformulcB,  the 
applications  of  which  are  to  be  made  elsewhere. 

Positive  and  negative  signs  in  trigonometry, 

192.  Before  entering  on  a  particular  consideration  of  the 
algebraic  expressions  which  are  produced  by  combinations  of 
the  several  trigonometrical  lines,  it  will  be  necessary  to  attend 
to  the  positive  and  negative  signs  in  the  different  quarters  of 
the  circle.  The  sines,  tangents,  &tc.  in  the  tables,  are  calcu- 
lated for  a  single  quadrant  only.  But  these  are  made  to  an- 
swer for  the  whole  circle.  For  they  are  of  the  same  length 
in  each  of  the  four  quadrants.  (Art.  90.)  Some  of  them, 
however,  are  ^Oiziit'e;  while  others  2iXe  negative.  In  alge- 
braic processes,  this  distiction  must  not  be  neglected. 

193.  For  the  purpose  of  tracing  the  changes  of  the  signs, 
in  different  parts  of  the  circle,  let  it  be  supposed  that  a  straight 
line  CT  (Fig.  36.)  is  fixed  at  one  end  C,  while  the  other  end 
is  carried  round,  like  a  rod  moving  on  a  pivot;  so  that  the 
point  S  shall  describe  the  circle  ABDH.  If  the  two  diame- 
ters AD  and  BH  be  perpendicular  to  each  other,  they  will 
divide  the  circle  into  quadrants. 

*  Euler's  Analysis  of  Infinites,  Hutton's  Mathematics,  Lacroix's  Differen- 
tial Calculus,  Mansfield's  Essays,  Legendre's,  Lacroix's,  Playfair's.Cagnoli's, 
and  Woodhouse's  Trigonometry. 

15 


106  TRlGONOxMETRICAL 

194.  ]n  (he  Jirst  guadrant  AB^  the  sine,  cosine,  tangent, 
&ic.  are  considered  all  positive.  In  the  second  quadrant  BD, 
the  sine  PS'  conunoes positive;  because  it  is  still  on  the  up- 
per side  of  the  diameter  AD,  from  which  it  is  measured. 
But  the  cosine,  which  is  measured  from  BH,  becomes  nega- 
tive, as  soon  as  it  changes  from  the  right  to  the  left  of  this 
line.  (Alg.  507.)  In  the  third  quadrant,  the  sine  becomes 
negative,  by  changing  from  the  upper  side  to  the  under  side 
of  DA.  The  cosine  continues  negative,  being  still  on  the  left 
of  BH.  In  ihejourth  quadrant,  the  sine  continues  negative. 
But  the  cosine  becomes  positive,  by  passing  to  the  right  of 
BH. 

195.  The  signs  of  the  tangents  and  secants  m^y  be  derived 
from  those  of  the  sines  and  cosines.     The  relations  of  these 
several  lines  to  each  other  must  be  such,  that  a  uniform  meth- 
od of  calculation  may  extend   through  the  different  quad- 
ants. 

In  the  first  quadrant,  (Art.  93.  Propor.  1.) 

T>  •       ,       •     rn         RXsin 

K  :  cos!  :tan  :  sin,  that  is,  ran= 

'  '  cos 

The  sign  of  the  quotient  is  determined  from  the  signs  of 
the  divisor  and  dividend.  (Alg.  123.)  The  radius  is  consid- 
ered as  always  positive.  If  then  the  sine  and  cosine  be  both 
positive,  or  both  negative,  the  tangent  will  be  positive.  But 
if  one  of  these  be  positive,  while  the  other  is  negative,  the 
tangent  will  be  negative. 

Now,  by  the  preceding  article. 

In  the  2d  quadrant,  the  sine  is  positive,  and  the  cosine  ne- 
gative. 

The  tangent  must  therefore  be  negative. 
In  the  3d  quadrant,  the  sine  and  cosine  are  both  negative. 

The  tangent  must  therefore  be  positive. 
In  the  4th  quadrant,  the  sine  is  negative,  and  the  cosine 
positive. 

The  tangent  must  therefore  be  negative. 

196.  By  the  9th,  3d,  and  6th  proportions  in  art.  93. 
1.  Tail  :  R:  :R  :  cot,  that  is,  Cot=  —  • 


ANALYSIS.  107 

Therefore,  as  radius  is  uniformly  positive,  the  cotangent 
must  have  the  same  sign  as  the  tangent. 

2.  Cos  :  R:  :R  :  sec.  that  is,  Sec= —  • 

The  secanf,  therefore,  must  have  the  same  sign  as  the  cosine. 

R2 

3.  Sin  :  R:  :R  :  cosec,  that  is,  Cosec  =  — r-  • 

sm 

The  cosecant,  therefore,  must  have  the  same  sign  as  the 
sine. 

The  versed  sine,  as  it  is  measured  from  A,  in  one  direction 
only,  is  invariably  positive. 

197.  The  tmgent  AT  (Fig.  36.)  increases,  as  the  arc  ex- 
tends from  A  towards  B.  See  also  Fig.  11.  Near  B  the  in- 
crease is  very  rapid ;  and  when  the  difference  between  the 
arc  and  90^,  is  less  than  any  assignable  quantity,  the  tangent 
is  ^rea/er  than  any  assignable  quantity,  and  is  said  to  he  infi- 
nite, (Alg.  447.)  If  the  arc  is  exactly  90  degrees,  it  lias, 
strictly  speaking,  no  tangent.  For  a  tangent  is  a  line,  drawn 
perpendicular  to  the  diameter  which  passes  through  one  end 
of  the  arc,  and  extended  till  it  meets  a  line  proceeding  from 
the  centre,  through  the  other  end.  (Art.  84.)  But  if  the  arc 
is  90  degrees,  as  AB,  (Fig.  36.)  the  angle  ACB  is  a  right  an- 
gle, and  therefore  AT  is  parallel  to  CB;  so  that,  if  these 
lines  be  extended  ever  so  far,  they  can  never  meet.  Still,  as 
an  arc  infinitely  near  to  90°  has  a  tangent  infinitely  great,  it 
is  frequently  said;  in  concise  terms,  that  the  tangent  of  90°  is 
infinite. 

In  the  second  quadrant,  the  tangent  is,  at  first,  infinitely 
great,  and  gradually  diminishes,  till  at  D  it  is  reduced  to 
nothing.  In  the  third  quadrant  it  increases  again,  becomes 
infinite  near  H,  and  is  reduced  to  nothing  at  A. 

The  cotangent  is  inversely  as  the  tangent.  It  is  therefore 
nothing  at  B  and  H,  (Fig.  36.)  and  infinite  near  A  and  D. 

198.  The  secant  increases  with  the  tangent,  through  the 
first  quadrant,  and  becomes  infinite  near  B  ;  it  then  dimin- 
ishes, in  the  second  quadrant,  till  at  D  it  is  equal  to  the  radius 
CD.  In  the  third  quadrant  it  increases  again,  becomes  infi- 
nite near  H,  after  which  it  diminishes,  till  it  becomes  equal 
to  radius. 

The  cosecant  decreases,  as  the  secant  increases,  and  v.  v. 
It  is  therefore  equal  to  radius  ;»t  B  and  H,  and  infinite  near 
A  and  D. 


108  TRIGONOMETRICAL 

199.  The  sine  increases  through  the  first  quadrant,  till  at 
B  (Fig.  36.)  it  is  equal  to  radius.  See  also  Fig.  13.  It  then 
diminishes,  and  is  reduced  to  nothing  at  D.  In  the  third  quad- 
rant, it  increases  again,  becomes  equal  to  radius  at  H,  and  is 
reduced  to  nothing  at  A. 

The  cositie  decreases  through  the  first  quadrant,  and  is  re- 
duced to  nothing  at  B.  In  the  second  quadrant,  it  increases, 
till  it  becomes  equal  to  radius  at  D.  It  then  diminishes  again, 
is  reduced  to  nothing  at  H,  and  afterwards  increases  till  it  be- 
comes equal  to  radius  at  A. 

In  all  these  cases,  the  arc  is  supposed  to  begin  at  A,  and  to 
extend  round  in  the  direction  of  BDH. 

200.  The  sine  and  cosine  vary  from  nothing  to  radius, 
which  they  never  exceed.  The  secant  and  cosecant  are  never 
less  than  radius,  but  may  be  greater  than  any  given  length. 
The  tangent  and  cotangent  have  every  value  from  nothing  to 
infinity.  Each  of  these  lines,  after  reaching  its  greatest  limit, 
begins  to  decrease  ;  and  as  soon  as  it  arrives  at  its  least  limit, 
begins  to  increase.  Thus  the  sine  begins  to  decrease,  after 
becoming  equal  to  radius,  which  is  its  greatest  limit.  But  the 
secant  begins  to  increase  after  becoming  equal  to  radius,  which 
is  its  least  limit. 

201 V  The  substance  of  several  of  the  preceding  articles  is 
comprised  in  the  following  tables.  The  first  shows  the  signs 
of  the  trigonometrical  lines,  in  each  of  the  quadrants  of  the 
circle.  The  other  gives  the  values  of  these  lines,  at  the  ex- 
tremity of  each  quadrant. 


Quadrant  1st 

2d 

3d 

4th 

Sine  and  cosecant 

+ 

+ 

— 

— 

Cosine  and  secant 

+ 

— 

— 

4- 

Tangent  and 

cotangent 

+ 

— 

-f 

0°       90° 

180° 

270° 

360° 

Sine 

0         r 

0 

r 

0 

Cosine 

r         0 

r 

0 

r 

Tangent 

0            CD 

0 

CD 

0 

Cotangent 

CD             0 

CD 

0 

CO 

Secant 

r         CO 

r 

CO 

r 

Cosecant 

CO        r 

CO 

r 

op 

Here  ris  put  for  radius,  and  co   for  infinite. 

202.  By  comparing  these  two  tables,  it  will  be  seen,  that 
each  of  the  trigonometrical  lines  changes  from  positive  to  ne- 
gative, or  from  negative  to  positive,  in  that  part  of  the  circle 


ANALYSIS.  lOD 

m  which  the  line  is  either  nothing  or  infinite.  Thus  the  tan- 
gent changes  from  positive  to  negative,  in  passing  from  the 
first  quadrant  to  the  second,  through  the  place  where  it  is 
infinite.  It  becomes  positive  again,  in  passing  from  the 
second  quadrant  to  the  third,  through  the  point  in  which  it  is 
nothing. 

203.  There  can  be  no  more  than  360  degrees  in  any  cir- 
cle. But  a  body  may  have  a  number  of  successive  revolu- 
lioBS  in  the  same  circle ;  as  the  earth  moves  round  tlie  suuj 
nearly  in  the  same  orbit,  year  after  year.  In  astronomical 
calculations,  it  is  frequently  necessary  to  add  together  parts 
of  different  revolutions.  The  sum  may  be  more  than  360*^ 
But  a  body  which  has  made  more  than  a  complete  revolution 
in  a  circle,  is  only  brought  back  to  a  point  which  it  had  pass- 
ed over  before.  So  the  sine,  tangent,  &c.  of  an  arc  great- 
er than  SGO"^,  is  the  same  as  the  sine,  tangent,  &;c.  of  some 
arc  less  than  360'.  If  an  entire  circumference,  or  a  number 
of  circumferences  be  added  to  any  arc,  it  will  terminate 
in  the  same  point  as  before.  So  that,  if  C  be  put  for 
a  whole  circumference,  or  360°,  and  x  be  any  arc  whatever: 

sin  x=sin  (C  +  x)  =  sin  (2C-f--i)=sin  (3C  +  a:),  &;c. 
tan  x  =  tan(C-h.T)=tan  (2C+T)  =  tan  (3C  +  x),  &c. 

204.  It  is  evident  also,  that,  in  a  number  of  successive  rev 
olutions,  in  the  same  circle; 

The  first  quadrant  must  coincide  with  the  5ih,  9th,  13th,  17th. 
The  second,  with  the  6th,  10th,  14th,  18th,  &;c. 

The  third,  with  the  7th,  11th,  15th,  19th,  &:c. 

The  fourth,  with  the  Sth,  12ih,  16th,  20th,  &c. 

205.  If  an  arc  extending  in  a  certain  direction  from  a  given 
point,  be  considered /?05tYj're;  an  arc  extending  from  the  same 
point,  in  an  opposite  direction,  is  to  be  considered  negative. 
(Alg.  507.)  Thus,  if  the  arc  extending  from  A  to  S  (Fig.  36.) 
be  positive;  an  arc  extending  from  A  to  S'"  will  be  negative. 
The  latter  will  not  terminate  in  the  same  quadrant  as  the  oth- 
er; and  the  sines  of  the  tabular  lines  must  be  accommodated 
to  this  circumstance.  Thus  the  sine  of  AS  will  be  positive, 
while  that  of  AS'"  will  be  negative.  (Art.  194.)  When  a 
greater  arc  is  subtracted  from  a  less,  if  the  latter  be  positive* 
ihe  remainder  must  be  negative.  (Alg.  58,  9.) 

Trigonometrical  Formula:. 

206.  From  the  view  which  has  here  been  taken  of  the 
changes  in  the  trigonometrical  lines,  it  will  be  easy  to  see,  in 


110  TR1G0N03IETKICAL 

what  parts  ofthe  circle  each  of  them  increases  or  decreases. 
But  this  does  not  determine  their  exact  values,  except  at  the 
extremities  of  the  several  quadrants.  In  the  analytical  in- 
vestigations which  are  carried  on  by  means  of  these  lines,  it 
is  necessary  to  calculate  the  chanf;es  produced  in  them,  by  a 
given  increase  or  diminution  of  the  arcs  to  which  they  be- 
long, 'n  this  there  would  be  no  difficulty,  if  the  sines,  tan- 
gents, &1C.  were  proportioned  to  their  arcs.  But  this  is  far 
from  ooing  the  case.  If  an  arc  is  doubled,  its  sine  is  not  ex- 
actly doubled.  Neither  is  its  tangent  or  secant.  We  have 
to  inquire,  then,  in  what  manner,  the  sine,  tangent,  &ic.  of 
one  arc  may  be  obtained,  from  those  of  other  arcs  already 
known. 

The  problem  on  which  almost  (he  whole  of  this  branch  of 
analysis  depends,  consists  in  deriving,  from  the  sines  and  co- 
bines  of  two  given  arcs,  expressions  for  the  sine  and  cosine 
of  t'leir  swn  and  difference.  For,  by  addition  and  subtrac- 
tion, a  few  arcs  may  be  so  combined  and  varied,  as  to  pro- 
duce others  of  almost  every  dimension.  And  the  expres- 
sions for  the  tangents  and  secants  may  be  deduced  from  those 
of  the  sines  and  cosines. 

Expressions  for  the  sine  and  cosine  of  the  sum  o/jc/ differ- 
ence of  arcs, 

207.  Let  rt  =  AH,  the  greater  of  the  given  arcs, 
And  6  =  HL  =  MD,  the  less.  (Fig.  37.) 

I'hen  «-f  i  =  AU-|-HL=AL,  the  swn  of  the  two  arcs, 
And    a -b=^All--HD=^ADythe\Y  difference. 

Draw  the  chord  DL,  and  the  radius  CII,  which  may  be 
represented  by  R.  As  DM  is,  by  construction,  equal  to  HL; 
DQ  is  equal  to  QL,  and  therefore  DL  is  perpendicular  to 
CH.  (Euc.  ^.  3.)  Draw  DO,  [IN,  QP,  and  LM.  each  per- 
pendicular lo  AC  ;  and  DS  and  QB  parallel  to  AC. 

From  the  definitions  of  the  sine  and  cosine,  (Art.  82,  0,) 
>J  is  evident,  that 

fof  AH,  that  is,  .9m  a=HN, 

'""^    "''^,  ofAL,i.e.sin(a-f6)  =  LM, 
lof  AD,i.o.?in(«-^)x=DO. 


ANALYSIS.  1 1 1 

fof  AH,  that  is,  cos  a=CN, 
r,„  .        ]  of  HL,  that  is,  cos  6=CQ, 

J  he  cos.ne  <;^  of  AL,  i.e.cosr^+M^CM, 

|^ofAD,i.e.cos(a-6)=CO. 

The  triangle  CHN  is  obviously  similar  to  CQP  ;  and  it  i^ 
also  similar  to  BLQ,  because  the  sides  of  the  one  are  per- 
pendicular to  those  of  the  other,  each  to  each.  We  have, 
then, 

1.  CH  :  CQ:  :IJN  :  QP,  that  is,  R  :  cos  b:  [sin  a  :  QP. 

2.  CH  :  QL:  :CN  :  BL,  R  :  sin  bwcos  a  :  BL, 

3.  CH  :CQ::CN  :  CP,  R  x  cos  bWcos  a  :CP. 

4.  CH  :  QL:  :HN  :  QB,  R  :  sin  br.sin  a  :  QB, 

Converting  each  of  these  proportions  into  an  equation  ; 
si7i  a  cos  b*  cos  a  cos  b 

I-  Qp=     R  3.  cp= — n — 

sin  b  cos  a                                       sin  a  sin  b 
2.    BL=       R  4.     QB=        ^ 


QP4-BL 


Then  adding  the  first  and  second. 
sin  a  cos  b~^sin  b  cos  a 


QP-BL= 


R 

Subtracting  the  second  from  the  first. 
sin  a  cos  b — sin  b  cos  a 


CP-QB 


CP4-QB: 


R 

Subtracting  the  fourth  from  the  third. 
cos  a  cos  b — sin  a  sin  b 


R 
Adding  the  third  and  fourth. 

cos  a  cos  b-\'sin  a  sin  h 


put 


R 

In  these  formulae,  the  sign  of  multiplication  is  omitted ;  stn  a  cos  b  bejin» 
for  sin  a  vi  cos  b,  that  i?  the  product  of  the  sine  of  a  into  thp  rr>smp  nf  '. 


1 12  TRIGONOMETRICAL 

But  it  will  be  seen,  from  the  figure,  that 

QP+BL=BM-f  BL=LM=5m  (a+b) 
QP-BL=QP-QS  =  D0=5UJ  (a-b) 
CP-QB=CP-PM==CM=c-^s  (a+b) 
CP+QB=CP+SD=CO=co5  (a-b) 

208.  If  then,  for  the  first  member  of  each  of  the  four  equa- 
tions above,  we  substitute  its  value,  we  shall  have, 

I.    o^^i^+h)-'"''  ^  cosb+sinb  cos  a 

H.     /        i-x     sin  a  cos  b  —  5m  b  cos  a 
.  sin  {a—o)=' 


III.  co^(a+6)  = 


R 

cos  a  cos  b  —  sin  a  sin  b 


R 

TIT         /        IS     cos  a  cos  b-^siyi  a  sinb 

ly,  cos  (a  —  b)= ^ 

R 

Or,  multipljing  both  sides  by  R, 

Rsin  {a-\-b)=sin  a  cos  b+sin  b  cos  a^ 
R  sin  {a — b)=sin  a  cos  b  — sin  b  cos  a 
R  {cos  {a-]rb)=cos  a  cos  b  —sin  a  sin  b 
R  cos  (a — b)  —  cos  a  cos  b-\-sina  sin  b 

That  is,  the  product  of  radius  and  the  sine  of  the  mm  of 
two  arcs,  is  equal  to  the  product  of  the  sine  of  the  first  arc 
into  the  cosine  of  the  second  -f  the  product  of  the  sine  of  the 
second  into  the  cosine  of  the  first. 

The  product  of  radius  and  the  sine  of  the  difference  of  two 
arcs,  is  equal  to  the  product  of  the  sine  of  the  first  arc  into 
the  cosine  of  the  second  — the  product  of  the  sine  of  the 
second  into  the  cosine  of  the  first. 

The  product  of  radius  and  the  cosine  of  the  sum  of  two 
arcs,  is  equal  to  the  product  of  the  cosines  of  the  arcs  —  the 
product  of  their  sines. 

The  product  of  radius  and  the  cosine  of  the  difference  of 
two  arcs,  is  equal  to  the  product  of  the  cosines  of  the  arcs  -f 
the  product  of  their  sines. 

These  four  equations  may  be  considered  as  fundamental 
propositions,  in  what  is  called  the  Arithmetic  of  Sines  and  Co- 
sines,  or  Trigonometrical  Analysis* 


ANALYSIS.  113 

Expressions  for  the  sine  and  cosine  of  a  double  arc* 

209.  When  the  sine  and  cosine  of  any  arc  are  given,  it  is 
easy  to  derive  from  the  equations  in  the  preceding  article,  ex- 
pressions for  the  sine  and  cosine  of  double  that  arc.  As  the 
two  arcs  a  and  b  may  be  of  any  dimensions,  they  may  be  sup- 
posed to  be  equal.  Substituting,  then,  a  for  its  equal  6,  the 
tirst  and  the  third  of  the  four  preceding  equations  will  be- 
come, 

R  sin  (a-{-a)=5m  a  cos  a-^sia  a  cos  a 
R  cos  (a-f-a)=c05  a  cos  a  — sin  a  sin  a 

That  is,  writing  sin- a  for  the  square  of  the  sine  of  «,  and 
^'os'^a  for  the  square  of  the  cosine  of  «, 

I.  R  sin  2a  =  2s ill  a  cos  a 

II.  R  cos  2a=cos^a  — sin^a* 

Expressions  for  the  sine  and  cosine  of  half  a  given  arc* 

210.  The  arc  in  the  preceding  equations,  not  being  neces- 
sarily limited  to  any  particular  value,  may  be  half  a,  as  well 
as  a.     Substituting  then  |a  for  a,  we  have, 

R  sin  a:=2sin  -i-o;  cos  -^a 
R  cos  a=cos^  ^a  —  sin^  ^u 

Putting  the  sum  of  the  squares  of  the  sine  and  cosine  equal 
10  the  square  of  radius,  (Art.  94.)  and  inverting  the  members 
of  the  last  equation. 


cos  2ia-|-5m-|a=R* 
co5^Ja  — sm^  Ja=R  cos  a 


If  we  subtract  one  of  these  from  the  other,  the  terras  con- 
taining cos^^a  will  disappear  ;  and  if  we  add  them,  the  terms 
containing  siti^^a  will  disappear  :  therefore, 

2sin^^a=K^  — -R  cos  a 
3cos2|«=R2+Rco5G 


TI4  TRIGONOMETRICAL 

Dividing  by  J,  and  extracting  the  root  of  both  sides, 

U.cos^a=  ^/Ul'^-h^RXcos  a 
Expressions  for  the  sines  and  cosines  o/'multiple  ai'cs, 

211.  In  the  same  manner,  as  expressions  for  the  sine  and 
cosine  of  a  double  arc,  are  derived  from  the  equations  in  art. 
208 ;  expressions  for  the  sines  and  cosines  of  other  muUiple 
arcs  ma}^  be  obtained,  by  substituting  successively  2a,  3a, 
Sic.  for  h,  or  for  h  and  a  both.     Thus, 


r  R^m  3rt=R  sin{a+2a): 
-?  Ksin  4ff=R  sinla-\-^)- 
(^  R.fi?i  5a=R  sin{a-\-4a) 


'Rsin  3rt=R  sin(a+2a)==sin  a  cos  2a-{-sm  2a  cos  a 

i=sin  a  cos  3fl-{-sm  3a  cos  a 

■sin  a  cos  ^a-^-sin  4a  cos  a 

&LC. 


CR  cos  3a =R  co5(a-i-2a)=co5  a  cos  2a  — sin  a  sin  2a 
IT.    <R  cos  4a=R  cos{a-\-Qa)=^cos  a  cos3a  —  sina  sin  Sa 
^R  cos  5a=R  cos{a-\-4a)=^cos  a  cos  Aa  —  sin  a  sin  Aa 
&;c. 

Expressions  for  the  products  of  sines  and  cosines, 

212,  Expressions  for  the  product  of  sines  and  cosines  may 
be  obtained,  by  adding  and  subtracting  the  four  equations  in 
art.  208,  viz. 

R  sin{a'\-b)=sin  a  cos  b-\-sin  b  cos  a 
R  sin{a  —  b)=sin  a  cos  b  —  sin  b  cos  a 
R  cos{a-\-b)=cos  a  cos  h  —  sin  a  sin  b 
R  cosla  —  b)  =  cos  a  cos  b-\-sin  a  sin  b 

Adding  the  first  and  second, 
71  sin{a-\-b)-\-R  sin(a—b)=2  sin  a  cos  b 

Subtracting  the  second  from  the  first, 
R  sin{a-{-b)  —  R  sin{a  —  b)~2sin  b  cos  a 

Adding  the  third  and  fourth, 
H  cos{a  —  b)-{'R  cos(a'}-b)=2cos  a  cos  h 

Subtracting  the  third  from  the  fourth, 
R  cos{a-^h)'^R  cos(a'\-b)=:2sin  a  sin  b 


ANALYSIS.  115 

Inverting  the  members  of  each  of  these  equations,  and  di- 
viding by  2,  we  have, 

I.  sin  a  cos  h=-\R  sin{a-{-h)-\-\R  siii{a'--b) 

II.  sin  h  cos  a=^R  sin(a-{-b)  —  ^R  s{ii(a—b) 

III.  cos  a  cosh  =  }jR  cos{a  -b)-\-^Rcos{a-{-b) 

IV.  sin  a  sin  b  =  ^R  cos{a  —  b)  —^R  C05(a-f- J) 

213.  If  6  be  (aken  equal  to  a,  then  a-\'b=2a,  and  a  — Z»=0, 
the  sine  of  which  is  0  ;  (Art.  201.)  and  the  term  in  which  this 
is  n factor,  is  reduced  to  0.  (Alg.  112.)  But  the  cosine  of  G 
is  equal  to  radius,  so  that  i?  X cos  0=/?^.  Reducing,  then, 
the  preceding  equations, 

The  first  becomes  sin  a  cos  a  =  ^Rsin  2a 

The  third,  cos-a  =  ^R- -\^^R  cos  2a 

The  fourth,  sin~a  =  lR-  -  ^R  cos  2a 

214.  If  s  be  the  sum.  and  d  the  difference  of  two  arcs, 
l{s-\-d)  will  be  equal  to  the  greater,  and  ^{s  —  d)  to  the  less. 
(Art.  153.)  Substituting  then,  in  the  four  equations  in  art. 
212, 

5  for  a-\-b  ^-(s  +  c?)  for  a 

d  for  a  —  b,  j(.s  —  d)  for  b,  we  have, 

I.  sin  ^(s-{-d{cos  l[s  —  d)  =  lR  (sin  s-{-sin  d) 

II.  5171  \{s  —  d)cos  \{s-\-d)  =  ];R  (sin  s  —  sin  d) 

III.  cos  ^{s-{-d)cos  iis  —  d)  =  ~R  {cos  d-\rcos  s) 

IV.  5m  l{s-{'d)sin  ~{s^d)=-lR  (cos  d  —  cos  s^ 

Or,  making  R=l, 

I.  sin  (a-{-b)-\-sin(a  —  b)=2  sin  a  cos  h 

n.  5m  (a-\-b)  —sin(a~b)  —  2  sin  b  cos  a 

III.  cos  (a  —  b)-{-cos(a-\-b)=i2cosacosh 

IV.  cos  (a  —  b)  —cos(a-{-b)  =  2  sin  a  sin  h 

215.  If  radius  be  taken  equal  to  1,  the  two  first  equations 
in  art.  208,  are, 

sin(a-\-b)=6in,  a  cos  b-\-sin  h  cos  q 
sin(a  —  h)=sin  a  cos  b  —  5m  b  cos  a 

Multiplying  these  into  each  other, 
^infa-^h)  Xsin(a  —  h)=.tin^a  ros^h—.v'n^b  ros'^ct 


lie  TRIGONOMETRICAL 

But  by  art.  94,  if  radius  is  1, 
cos^h=:l—si7i^b,  and  co9-a=l—sin^a 

Substituting,  then,  for  cos^h  and  eos^a,  their  values,  multi- 
plying the  factors,  and  reducing  the  terms,  we  have, 

sin(a'\-b)Xsi72{a  —  h)=s{n^a—s{n^b 

Or,  because  the  difference  of  the  squares  of  two  quantities 
is  equal  to  the  product  of  their  sum  and  difference,  (Alg. 
235.) 

sin{a'-\-b)Xsin{a  —  b)=(sina'\'sin  b)  X  (siii  a  —sin  b) 

That  is,  the  product  of  the  sine  of  the  sum  of  two  arcs,  into 
the  sine  of  their  difference ;  is  equal  to  the  product  of  the  sum 
of  their  sines,  into  the  difference  of  their  sines. 

Expressions  for  the  tangents  of  arcs. 

21 G.  Expressions  for  the  tangents  of  arcs  may  be  derived 
from  those  already  obtained  for  the  sines  and  cosines.  By 
art.  93,  proportion  1st, 

R  \  tan',  'cos  l  sin 

R      cos  tan      sin  Rxsi 

•^^3t  is,  TTT^Tv.,  and  -5-=— ,  and  tan- 


sin 


tan     sin'  R      cos"*  cos 

r^,  .       ,     ^  sin(a-\-b) 

Thus  tan(a-\-b=^ ?--7Tr^' 

^     *  cos{a-\-o) 

If,  for  shi{a-\-b)  and  cos{a-]rb)  we  substitute  their  values,  as 
given  in  art.  208,  we  shall  have, 

R{sin  a  cos  b-^sin  b  cos  a) 

tan(a-\-b)= j : — r~* 

^         ^         cos  a  coso  —  sm  a  sm  0 

217.  Here  the  value  of  the  tangent  of  the  sum  of  two  arcs 
is  expressed,  in  terms  of  the  sines  and  cosines  of  the  arcs.  To 
exchange  these  for  terms  of  the  tangents  let  the  numera- 
tor and  denominator  of  the  second  member  of  the  equation 
be  both  divided  by  cos  a  cos  b.  This  will  not  alter  the  value 
of  the  fraction.  (Alg.  140.J 

The  numerator,  divided  by  cos  a  cos  b,  is 

R(sin  a  cos  b-^-sin  b  cos  a)  /sin  a     sin  b\ 

— ■ j =R  I + i)-=:tana-\-tano 

cos  a  cos  b  \cos  r     cos  0/ 


ANALYSIS.  jir 

And  the  denominator,  divided  by  cos  a  cos  b,  is 
OS  a  cos  b  —  sin  a  sin  b  sin  a     sin  b  tan  a     tan  h 


cos  a  cos  b  cos  a     cos  b  R  R 

tan  a-{-tan  b 
Therefore  ^«'K«+^)  =1^^^7^ 

1-         R^ 

The  denominator  of  the  fraction  may  be  cleared  of  the  di- 
visor i2-,  by  muhiplying  both  the  numerator  and  denomina- 
tor into  R^.  And  if  we  proceed  in  a  similar  manner  to  find 
the  tangent  ofa  —  b,  we  shall  have, 

R-(tan  a-rtan  b) 

218.  I.  tan{a-^b)=^-^- — ~ 

V         ^      H^  —tan  a  tan  b) 

R-(tan  a  —  tan  b) 
II.  t<'ni'i+h)=  R.+tanaUl^Tb) 
If  the  arcs  a  and  b  are  equal,  then  substituting  \a,  a,  2a,  oa. 
&G.  as  in  art.  210,  211, 

R^2tan^a) 
tan  a=tan(~,a-i'ia)=n:^ — : — tt 

R^2tan  a) 
tan  '^^=tan{a^a)  =  -^^-::j^^ 

R-(ian  a-rtan  2a)    « 
tanSa=tan{a+2a)=-jjr=J^---J^^^2^^  ^'■ 

219.  If  we  divide  the  first  of  the  equations  in  art.  214,  by 
the  second  ;  we  shall  have,  after  rejecting  hR-  from  the  nu- 
merator and  denominator,  (Alg.  140.) 

sin^(s-\-d)cos~{s  —  d)     sin  s-\-sin  d 

sin~(s  —  d)cosl{s-\-d)~~sin  s-  sin  d 

But  the  first  member  of  this  equation,  (Alg.  155,)  is  equal  to 

sin\{s+d)_cosi(s-d)^tanl{s'{-d)^ R 

cos^^(s^d)     sini (s-d)  R  tan  i{s-'d)  ^'^^'^    '' 

Therefore, 

sin  s-\-sin  d_  tan  2(^4-^) 

sin  5  — S7n  d      tan  ^(s  —  d) 


i  IC  TRIGONOMETRICAL 

220.  According  to  the  notation  in  art.  214,  s  stands  for  the 
sum  of  two  arcs,  and  d  for  their  difference.  But  it  is  evident 
that  arcs  may  be  taken,  whose  sum  shall  be  equal  to  any  arc 
a,  and  whose  difference  shall  be  equal  to  any  arc  6,  provided 
that  a  be  greater  than  b.  Substituting  then,  in  the  preceding 
equation,  a  f  jr  s,  and  b  for  d, 

sin  a-^-sin  b __taa  \{a-\-b) 
sin  a  —sin  b     tan  ^(a  —  b)  ' 

sin  a-^-sin  b  :  sin  a— sin  b'.'.tan  ~{a-\-b)  \  tan  \{a  —  b,) 

That  is,  The  sum  of  the  sines  of  two  arcs  or  angles,  is  to  the 
difference  of  those  sines;  as  the  tangent  of  half  the  sum  of  the 
arcs  or  angle^  to  the  tangent  of  half  their  difference* 

By  art.  143,  the  sides  of  triangles  are  as  the  sines  of  their 
opposite  angles.  It  follows,  therefore,  from  the  preceding 
proposition,  (Alg,  389.)  that  the  sum  of  any  two  sides  of  a 
triangle,  is  to  their  difference  ;  as  the  tangent  of  half  the  sum 
of  the  opposite  angles,  to  the  tangent  of  half  their  difference. 

This  is  the  second  theorem  applied  to  the  solution  of  ob- 
lique angled  triangles,  which  was  geo/7ieinca%  demonstrated 
in  art.  144. 

Expressions  for  the  cotangents  may  be  obtained  bv  putting 

cot  =  ^^(Art.  93.) 

__3l—  R' -tan  ft  tan  6  ^^^^  ^is.) 
Thus  cot  («+6)=tan  {a'\-b)=  tan  «-f  tan  h 

Substituting  cot  a  ^^r  tan  ft,  and  ^ot  A  ^or  tan  //, 

R2_5i! ?:!_ 

cot  ft  ^cot  i) 


cot(ft-f^)=         R2  R2 

cotff'^cot6 

Multiplying  both  the  numerator  and  denominator  by  cot  u 
cot  6,  dividing  by  R-,  and  proceeding  in  the  same  manner, 
for  cot  (a—b)  we  have, 

cot  ft  cot  6-R- 
T;  rot  (ft4-6)=     cot^4-cota 


ANALYSIS.  119 

cot  a  cot  6-{-R= 
II.  cot  {a  -  h)—    cot  6— cot  « 

220.  h.  By  comparing  the  expressions  for  the  sines,  and 
cosines,  with  those  for  the  tangents  and  cotangents,  a  great 
variety  of  formulae  may  be  obtained.  Thus  the  tangent  of 
the  sum  or  the  difference  of  two  arcs,  may  be  expressed  in 
terms  of  the  cotangent. 

Puttjng  radius  =1,  we  have  (Art.  93,  220.) 

1  cot6-|-cot  a 

I.  tan  («+i)=cot  (a+i)=^^U^t'A^l 

I  cot  6  — cot  a 

II.  tan  (a  — 6)= — -r — 7t= — : rrvT 

^         ^     cot(a  — 6)     cotacot6-fl 

By  art.  208, 

sin  (ff+6)        sin  a  cos  6-{-sin  h  cos  a 
sin  (a —  6)  ~  sin  «  cos  6  —  sin  b  cos  a 

Dividing  the  last  member  of  the  equation,  in  the  fust  place 
by  cos  a  cos  6,  as  in  art.  217,  and  then  by  sin  a  sin  b,  we 
have 

sin  («+6)       tan  «4-tan /•       cot^-f-cotw 
sin  {a  —  b)  "~"  tan  a  —  tan  6  ~  cot  b  —  cot  a 

In  a  similar  manner,  dividing  the  expressions  for  the  co- 
sines, in  the  first  place  by  sin  b  cos  or,  and  then  by  sin  a  cos 
h^  we  obtain 

cos  {a+b)  _^  cot  b  —  tan  a       cot  a  —  tan  b 
cos  {a  —  h)  ~"  cot  />-f-tan  a  "^  cot  c+tan  b 

Dividing  the  numerator  and  denominator  of  the  expres- 
sion for  the  tangent  of  a,  (Art.  210.)  by  tan  \a,  we  have 


tana       ^ot  4«-tan  ^a 

These  formulas  may  be  multiplied  almost  indefinitely,  by 
•mbining  the  expression?  for  the  sines,  tangents,  &c.     The 


120  TRIGONOMETRICAL 

following  are  put  down  without  demonstrations,  tor  the  exer 
cise  of  the  student. 


1  —cos  a 

tan  ^a=cot|(z  —  2  cot  a.  tan  -5-a~     sin  a 

sin  a  I  —cos  a 

tan  ia=7-r— "~  tan"|a=7-; 

l-t-cos«  2       i-j-cosff 

2  tan  |fl  ^  1— tan  ^J« 

5in  C5  —  i4.tan2|«:  cos  a  —  i-^tanH^ 

cot  |a  — tan  \a  -                    2 


cos  a     cot  ifl+tan  »«      sm  a  —cot  |a+tan  ^« 
1  1 


sin  a—cot  |«  -  cot  a  sin  a     cot  «-|-tan  la 


Expression  for  the  area  of  a  triangle,  in  terms  of  the  sides. 

221.  Let  the  sides  of  the  triangle  ABC  (Fig.  23.)  be  ex- 
pressed by  a,  b,  and  c,  the  perpendicular  CD  by  p,  the  seg- 
ment AD  by  dy  and  the  area  by  S. 

Thena2=62-f.c=-2ctZ,  (Euc.  13.  2.) 

Transposing  and  dividing  by  2c 


d= 


6^-fc^-Q"  (6M-c2--a3)2     (Alg. 

2c        •      Therefore  rf^=  iT^         •      [223.) 

By  Euc.  47.  1 ,  p2  =^2  _ ^2  =^3  _:^ _ — l- 

Reducingthe  fraction,  (Alg.  150.)  and  extracting  the  root 
o^  both  sides. 


ANALYSIS.  121 


3 


This  gives  the  length  of  the  perpendicular,  in  terms  of  the 
sides  of  the  triangle.  But  the  area  is  equal  to  the  -product 
of  the  base  into  half  the  perpendicular  height.     (Alg.  518.) 

that  is,  . 

^=^\cp  =  W\b'c^  -  {b'  -\'C'-a^y 

Here  we  have  an  expression  for  the  area,  in  terms  of  the 
sides.  But  this  may  be  reduced  to  a  form  much  better 
adaj^ed  to  arithmetical  computation.  It  will  be  seen,  that 
the  quantities  4h-c-,  and  {b- -\-c''  -a^Y  are  both  squares  ; 
and  that  the  whole  expression  under  the  radical  sign  is  the 
difference  of  these  squares.  But  the  difference  of  two  squares 
is  equal  to  the  product  of  the  sum  and  difference  of  their 
roots.  (Alg.  233.)  Therefore  46-c=^ -(6^ -fc' -a')'  maybe 
resolved  into  the  two  factors, 


(  26c4-(^'  4-c2  -a2)  which  is  equal  to  (^-hc)^  -a» 
(  26c  — (6=^-1- c^"  -a^)  which  is  equal  toa^—^^^  —  c)^ 

Each  of  these  also,  as  will  be  seen  in  the  expressions  on 
the  right,  is  the  difference  of  two  squares  ;  and  may,  on  the 
same  principle,  he  resolved  into  factors,  so  that, 

Ub-\-cY-a^-=ib'\-c-\-a)X{b-\-c-a) 
la^-{b-  c)'  =(a-f-6-c)  X(«— 6+c) 

Substituting,  then,  these  four  factors,  in  the  place  of  the 
quantity  which  has  been  resolved  into  them,  we  have, 

S  =  iv/(6-{-c-f  a)  X  (6+c-a)  X(a+6-c)  X(a-6+c) 


*  The  expression  for  the  perpendicular  is  the  same,  when  one  of  the  angples 
13  obtuse,  as  in  Fig.  24.     Let  AD=:d. 

Then  a^  =6^  -f  c^  +2crf.  (Euc.  12.  2.)  And£?=  Yc 


(-^2-c2-j-a2)2  (^2^c2_a2)2 

4c2  =  ^ 


Therefore  (^^  = -'2 =  73 ^rAl.219,) 


And/;—  ^7  as  above 


17 


122  TRIGONOMETRICAL  ANALYSIS. 

Here  it  will  be  observed,  that  all  the  three  sides,  a,  6,  and 
c,  are  in  each  of  these  factors. 

Let  h=l{a-{-b'\'c)  half  the  sum  of  the  sides.     Then 


S=s/hy.{h'-a)X{h-b)x{h''c) 

222.  For  finding  the  area  of  a  triangle,  then,  when  the 
three  sides  are  given,  we  have  this  general  rule  ; 

From  half  the  sum  of  the  sides,  subtract  each  side  severally  / 
multiply  together  the  half  sum  and  the  three  remainders  ;  and 
extract  the  square  root  of  the  product.  H 


SEcnoN  vm 


COMPUTATION  OF  THE  CANON. 


Art  2^3  HPHE  trigonometrical  canon  is  a  set  of  tables 
-*-  containing  the  sines,  cosines,  tangents,  fee- 
to  every  degree  and  minute  of  the  quadrant.  In  the  com- 
putation of  these  tables,  it  is  common  to  find,  in  the  first 
place,  the  sine  and  cosine  of  one  minute  ;  and  then,  by  suc- 
cessive additions  and  multiplications,  the  sines,  cosines,  Sic* 
of  the  larger  arcs.  For  this  purpose,  it  will  be  proper  to 
begin  with  an  arc,  whose  sine  or  cosine  is  a  known  portion 
of  the  radius.  The  cosine  of  60°  is  equal  to  half  radius. 
(Art.  96.  Cor.)  A  formula  has  been  given,  (Art.  210.)  by 
which,  when  the  cosine  of  an  arc  is  known,  the  cosine  of 
half  ihdii  arc  may  be  obtained. 


By  successive  bisect! 

ions  of  60°,  we  have  the  arcv 

30° 

0°   28'    7'     30'" 

15° 

0     14     3      45 

T"  30' 

0      7      1       52  30 

3°  45' 

0       3  30      56   15 

1°  52'  30' 

0      1   45      28     7     30 

0°56' 

15" 

0       0'  52"     44'"  3""  45"'" 

By 

formula  II,  art.  210, 

cos  ia  =  v^iR2-fiRxco5  a 
If  the  radius  be  1,  and  if  a=60°,  6=30*^,  c=15°,  &c. ;  then 


cos6=cos  |a=v/i-f-iX  1=0.8660254 
cos  c=cos  ^b  =  y/^-\-lcos  6=0.9659258 


cos  d=cos  ^c=\/i-f-icos  c=0.99 14449 
cos  e=co5  ^</=\/|-f-|co5c^=s0.9978589 


124  COMPUTATION 

Proceeding  in  this  manner,  by  repeated  extractions  of  the 
square  root,  we  shall  find  the  cosine  of 

Qo  0'  52"  44'"  3""  45""'  to  be  0.99999996732 


And  the  sine  (Art.  94.)  =  ^/ 1 —C052  =0.00025566346 

This,  however,  does  not  give  the  sine  o{  one  minute  exact- 
ly. The  arc  is  a  little  less  than  a  minute.  But  the  ratio  of 
very  small  arcs  to  each  other,  is  so  nearly  equal  to  the  ratio 
of  their  sines,  that  one  may  be  taken  for  the  other,  without 
sensible  error.  Now  the  circumference  of  a  circle  is  divided 
into  21600  parts,  for  the  arc  of  1' ;  and  into  24576,  for  the 
arc  of  0°  0'  52"  44"'  3""  45""' 

Therefore, 

21600  :  24576:: 0.00025566346  :  0.0002908882. 

which  is  the  sine  of  I  minute  very  nearly.* 


And  the  cosine  =  v/ 1  -  sin^  =0-9999999577 

224.  Having  computed  the  sine  and  cosine  of  one  minute, 
we  may  proceed,  in  a  contrary  order,  to  find  the  sines  and 
cosines  of  larger  arcs. 

Making  radius  =1,  and  adding  the  two  first  equations  in 
art.  208,  we  have 

si7i{a-\-b)'{-sm(a  —  b)=2sin  a  cos  b 

Adding  the  third  and  fourth, 

cos{a-\'b)-\-cos(a  —  b)  =  2cos  a  cos  b 

Transposing  sin{a^b)  and  cos{a  —  b) 

1.    sin(a-{-h)=^sin  a  cos  b—  sm(a  —  b) 
n.    cos{a-{'b)  =  2cos  a  cds  b  —  cos{a —b) 

If  we  put  6  =  1',  and  c=l',  2',  3',  &c.  successively,  we 
hhali  have  expressions  for  the  sines  and  cosines  of  a  series  of 
arcs  increasing  regularly  by  one  minute.     Thus, 

•See note  H. 


OF  THE  CANOJS.  ]2r. 

6m(r4-l')  =  267Vil'Xcosl'  —  Wn  0=0.0005817764, 
sin{2' -{-V)=2sin2'  Xcos  V-^sin  l'=0  0008726645, 
.nn{3f  -{-r)  =  2sin  3'  Xcos  V  -  sin2'  =:0,00ll63552e. 
&c.  &c. 

(.05(l'4-l')=2cos  rXcos  l'-co5  0=09999998308 
ro5(2'-|-l')  =  2co5  2' Xco5l'  — co^  l'=0.9999996192 
<:o5(.3'-i-l')=2co5  3' Xcosl'  — C05 2'=0.9999993230 
&c.  &c. 

The  constant  ntiultiplier  here,  cos  V  is  0.9999999577,  which 
is  equal  to  1—0.0000000423. 

225.  Calculating,  in  this  manner,  the  sines  and  cosines 
from  1  minute  up  to  30  degrees,  we  shall  have  also  the  sines 
and  cosines  from  60°  to  90°.  For  the  sines  of  arcs  between 
0°  and  30°,  are  the  cosines  of  arcs  between  60°  and  90°.  And 
the  cosines  of  arcs  between  0°  and  30°,  are  the  sities  of  arcs 
between  60  and  90°.  (Art.  104.) 

226.  For  the  interval  between  30°  and  60°,  the  sines  and 
cosines  may  be  obtained  by  subtraction  merely.  As  twice 
the  sine  of  30°  is  equal  to  radius  ;  (Art.  96.)  by  making  a= 
30°,  the  equation  marked  I,  in  article  224  will  become 

5m(30°  +  6)=cos  6-5m(30°-6) 

And  putting  6  =  1',  2',  3',ke,  successively. 

sin{30^  r)=cos  l'-52>i(29°  59') 
(30°  2')=cos  2'  -  sin(29°  58') 
(30°  3')=cos  3'  —  sin{29°  51') 
&IC.  &;c. 

If  the  517165  be  calculated  from  30°  to  60°,  the  cosinea  wjii 
also  be  obtained.  For  the  sines  of  arcs  between  30°  and  45°, 
are  the  cosines  of  arcs  between  45°  and  60*.  And  the  sines 
of  arcs  between  45°  and  60°,  are  the  cosines  of  arcs  between 
30°  and  45°.*  (Art.  96. 

227.  By  the  methods  which  have  here  been  explained,  the 
natural  sines  and  cosines  are  found. 

The  logarithms  of  these,  10  being  in  each  instance  added 
to  the  index,  will  be  the  artificial  sines  and  cosines,  by  which 
trigonometrical  calculations  are  commonly  made,  (Art. 
102,  3.) 

228.  The  tangents,  cotangents,  secants,  and  cosecants,  art 
easily  derived  from  the  sines  and  cosines.     By  art.  93, 

*  See  note  \. 


!6  COMPUTATION  OF  THE  CANON 

R  t  cos !  I  tan  *  sin  cos  t  Rl'.R  *  sec 

R  :  sin : '.  cot  :  cos  sin  t  R  '.<  R  I  cosec 

Therefore, 
R  X  sin  , ,  R- 

The  lanG;ent=— The  secant= 

°  cos  cos 

R  X  cos  ^  R* 

The  cotangent=— ^—  The  cobecant=^^-^ 

Or,  if  the  computations  are  made  by  logarithms, 

The  tangent=  lO-f  5m  — coi,         The  secant = 20— co5, 
The  cotangent=5ilO-{-co.9  — 5iw,     The  cosecant =20— 5m. 


SECTION  IX. 


PARTICULAR  SOLUTIONS  OF  TRIANGLES.* 


A  9*^1  \^^  triangle  whatever  may  be  solved,  by 
ART.  ^oi,  x\.  ^j^g  theorems  in  sections  IIL  IV.  But 
there  are  other  methods,  by  which,  in  certain  circumstances, 
the  calculations  are  rendered  more  expeditious,  or  more  accu- 
rate results  are  obtained. 

The  differences  in  the  sines  of  angles  near  90°,  and  in  the 
cosines  of  angles  near  0^,  are  so  small  as  to  leave  an  uncertain- 
ty of  several  seconds  in  the  result.  The  solutions  should  be 
varied,  so  as  to  avoid  finding  a  very  small  angle  by  its  cosine, 
or  one  near  90°  by  its  sine. 

The  differences  in  the  logarithmic  tangents  and  cotangents 
are  least  at  45°,  and  increase  towards  each  extremity  of  the 
quadrant.  In  no  part  of  it,  however,  are  they  very  small.  In 
the  tables  which  are  carried  to  7  places  of  decimals,  the  least 
difference  for  one  second  is  42.  Any  angle  may  be  found 
within  one  second,  by  its  tangent,  if  tables  are  used  which  are 
calculated  to  seconds. 

But  the  differences  in  the  logarithmic  sines  and  tangents, 
within  a  few  minutes  of  the  beginning  of  the  quadrant,  and  in 
cosines  and  tangents  within  a  few  minutes  of  90°,  though  they 
are  very  large,  are  too  unequal  to  allow  of  an  exact  determin- 
ation of  their  corresponding  angles,  by  taking  proportional 
parts  of  the  differences.  Very  small  angles  may  be  accurate- 
ly found,  from  their  sines  and  tangents,  by  the  rules  given  in 
a  note  at  the  end. t 

232.  The  following  formulae  may  be  applied  to  right  angled 
triangles,  to  obtain  accurate  resuUs,  by  finding  the  sine  or  tan- 
gent of  Aa//*an  arc,  instead  of  the  whole. 

In  the  triangle  ABC  (Fig.  20,  PI.  II.)  making  AC  radius, 

AC  :  AB::i  :  Cos  A. 

By  conversion,  (Alg.  369,  5.) 

AC:  AC-AB::l  :  l-CosA. 

*  Simson's,  Woodhouse'?.  and  Caj^oli's  Tngonometry.        t  See  Note  K. 


Ua  PARTICULAR  SOLUTIONIS 

TherefoFe, 

AC-AB 

^^      =1— Cos  A=2sin^iA.  (Art.  210.; 

Sin  ^  A=v^ 


2AC 

Again,  from  the  first  proportion,  adding  and  subtracting 
lerms,  (Al§.  ,iS9,  7.) 

AC+AB  :  AC-AB::l-l-CosA  :  l~Cos  A. 
Therefore, 
AC-AB     1-CosA  ,      „ 

AC  +  AB^TTC^^^^^^  ^^-  ^^'^''  ^'^-^ 
Or, 

AC-AB 

TaniA=v'-^_p^ 

233.  Sometimes,  instead  of  having  two  parts  of  a  right  an- 
gled triangle  given,  in  addition  to  the  right  angle;  we  have 
only  one  of  the  parts,  and  the  sum  or  difference  of  two  others. 
In  such  cases,  solutions  may  be  obtained  by  the  following 
proportions. 

By  the  preceding  formulae,  and  Art.  140,  141, 
AC-AB 

1.  Tan^  iA=^:cTAB 

2.  BC2=(AC-AB)(AC-fAB) 

Multiplying  these  together,  and  extracting  the  root,  we 
hare, 

TaniAxBC=AC-AB 

Therefore, 

I.  Tani  A  :  i::AC-AB  :  BC 

That  IS,  the  tangent  of  half  of  one  of  the  acute  angles,  is  to 
1,  as  the  difference  between  the  hypothenuse  and  the  side  at 
the  angle,  to  the  other  side. 

If,  instead  of  multiplying,  we  divide  the  first  equation  abovxj 
by  the  second,  we  have 

TanfA 1___ 

BC    ~AC+AB 


OF  TRIANGLES.  129 

Therefore, 

II.  1  :tan  ^A::AC-f  AB  :  BC 

Again,  in  the  triangle  ABC,  Fig.  20, 

AB  :  BC  :  l::tan  A 

Therefore, 

AB+BC  :  AB-BC::l  +  tan  A  :  1-tan  A 

Or, 

1    fori     A 

AB-f  BC  :  AB-BC : :  1  :  t-; t 

l-f  tan  A 

By  art.  218,  one  of  the  arcs  being  A,  and  the  other  45°,  the 
tangent  of  which  is  equal  to  radius,  we  have, 
_  1  — tan  A 

Tan  i'''-^)=rH^^ 

Therefore, 
III.  1  :  tan  (45°-A)::AB4-BC  :  AB-BC. 
That  is,  unity  is  to  the  tangent  of  the  difference  between 
45°  and  one  of  the  acute  angles  ;  as  the  sum  of  ihe  perpendic- 
ular sides  is  to  their  difference. 

Ex.  1.  In  a  right  angled  triangle,  if  the  difference  of  the 
hypothenuse  and  base  be  64  feet,  and  the  angle  at  the  base 
33°f,  what  is  the  length  of  the  perpendicular.^ 

Ans.  211. 

2.  If  the  sum  of  the  hypothenuse  and  base  be  185.3,  and 
the  angle  at  the  base  37°;  what  is  thje  perpendicular.'* 

Ans.  620. 

3.  Given  the  sum  of  the  base  and  perpendicular  128.4,  and 
the  angle  at  the  base  41°i,  to  find  the  sides. 

1  :  tan(45°—4Pi)::  128.4  :  8.4,  the  difference  of  the 
base  and  perpendicular.  Half  the  difference  added  to,  and 
subtracted  from,  the  half  sum,  gives  the  base  68.4,  and  the 
perpendicular  60. 

4.  Given  the  sum  of  the  hypothenuse  and  perpendicular 
83,  and  the  angle  at  the  perpendicular  40°,  to  find  the  base. 

5.  Given  the  difference  of  the  hypothenuse  and  perpendic- 
ular 16.5,  and  the  angle  at  the  perpendicular  37^  j,  to  find  the 
base. 

6.  Given  the  difference  of  the  base  and  perpendicular  35, 
and  the  ang^le  at  the  perpendicular  27°|,  to  find  the  sides. 

18 


130  PARTICULAR  SOLUTIONS 

234.  The  following  solutions  may  be  applied  to  the  third 
'And  fourth  cases  o(  oblique  angled  triangles  ;  in  one  of  which, 
two  sides  and  the  included  angle  are  given,  and  in  the  other, 
the  three  sides.     See  pages  87  and  88. 

Case  IU. 


In  astronomical  calculations,  it  :s  frequently  the  case,  that 
two  sides  of  a  triangle  are  given  by  their  logarithms.  By 
the  following  proposition,  the  necessity  of  finding  the  corres- 
ponding natural  numbers  is  avoided. 

l^HEOitEM  A.  In  any  plane  triangle,  oj  the  two  sides  which 
include  a  given  angle,  the  less  is  to  the  greater  ;  as  radius  to 
the  tangent  of  an  angle  greater  than  45°  : 

And  radius  is  to  the  tangent  of  the  excess  of  this  angle  above 
45®  ;  as  the  tangent  of  half  the  sum  of  the  opposite  angles,  to 
the  tangent  of  half  their  difference 

In  the  triangle  ABC,  (Fig.  39.)  let  the  sides  AC  and  AB, 
and  the  angle  A  be  given.  Through  A  draw  DH  perpendic- 
ular to  AC.  Make  AD  and  AF  each  equal  to  AC,  and  AH 
equal  to  AB.  And  let  HG  be  perpendicular  to  a  line  drawn 
from  C  through  F. 

Then  AC  :  AB::R  :  Tan  ACH 
And  R  :  Tan(ACH  -  45°) : :  Tani( ACB-f  B):Tani(ACB  -  B) 

Demonstration, 

In  the  right  angled  triangle  ACD,  as  the  acute  angles  are 
subtended  by  the  equal  sides  AC  and  AD,  each  is  45°.  For 
the  !^ame  reason,  the  acute  angles  in  the  triangle  CAF  are 
each  45°.  Therefore,  the  angle  DCF  is  a  right  angle,  the 
angles  GFH  and  GHF  are  each  45°,  and  the  line  GH  is  equal 
to  GF  and  parallel  to  DC. 

In  the  triangle  ACH,  if  AC  be  radius,  AH  which  is  equal 
to  AB  will  be  the  tangt  nt  of  ACH.     Therefore, 

AC  :  AB::R  :  Tan  ACH. 

In  the  triangle  CGIJ,  if  CG  be  radius,  GH  which  is  equal 
to  FG  will  be  the  tangent  of  HCG.     Therefore, 

R  :  Tan  (ACH-45°):  iCG  :  FG. 


OF  TRIANGLES.  131 

And,  as  GH  and  DC  are  parallel,  (Euc.  -2.  6.) 
CG  :  FG::DH  :  FH. 

But  DH  is,  by  construction,  equal  to  the  sum,  and  FH  to 
the  difference  of  AC  and  AB  And  by  theorem  II,  [Art.  144.] 
the  sum  of  the  sides  is  to  their  difference  ;  as  the  tangent  of 
half  the  sum  of  the  opposite  angles,  to  the  tangent  of  half 
their  difference.     Therefore, 

R  :  Tan  (ACH-45°):  iTan  ^(ACB-f  B)  :  Tan  ^(ACB^B) 

Ex.  In  the  triangle  ABC,  (Fig.  30,)  given  the  angle  A  = 
26°  14',  the  side  AC  =  39,  and  the  side  AB=53. 

AC        39  1.5910646   R  10. 

AB       53  1.7242759  Tan  3°  39' 9"  9.1823381 

R  10.  Tan  ^(B4-C)76«'53'  10.632^181 


Tan 53° 39' 9"  10. 1332 11  3  Tani(B^C)33°8'50"  9.8149562 


The  same  result  is  obtained  here,  as  by  theorem  II,  p.  75. 

To  find  the  required  side  in  this  third  case,  by  the  theo- 
rems in  section  IV,  it  is  necessary  to  find,  in  the  first  place, 
an  aiigle  opposite  one  of  the  given  sides.  But  the  required 
side  may  be  obtained,  in  a  different  way,  by  the  following 
proposition. 

Theorem  B.  Iu  a  plane  triangle,  twice  the  product  of  any 
two  sides,  is  to  the  differeiice  between  the  sum  of  the  squares  oj 
those  sides,  and  the  square  of  the  third  side,  as  radius  to  the 
§osine  of  the  angle  included  between  the  two  sides. 

Id  the  triangle  ABC,  (Fig.  23.)  whose  sides  are  a,  b,  and  c, 

2bc  :  62-f-c2-«2;:R  ;  Cos  A 

For  in  the  right  angled  triangle  ACD,      b  :  d'.'.R  :  Cos  A 
Multiplying  by  2c,  26c  :  2rfc:  :R  :  Cos  A 

But,  by  Euclid  13.  2,  2dc  =  b^ -{-c^ -a'' 

Therefore,  26c  :  b'' -^-c^ -a^  ::R  :  Cos  A. 

The  demonstration  is  the  same,  when  the  angle  A  is  obtuse, 
as  in  the  triangle  ABC,  (Fig.  24.)  except  that  a^  is   greater 


132 


PARTICULAR    SOLUTIONS 


than  b--{-c'  ;  ''Eiic.  12.  2.)   so  that  the  cosine  of  A  is  nega- 
tive. See  art    194. 

From  this  theorem  are  derived  expressions,  both  for  the 
sides  of  a  tr'angle,  and  for  the  cosines  of  the  angles.  Con- 
verting the  last  proportion  into  an  equation,  and  proceeding 
in  the  same  mannei  with  the  other  sides  and  angles,  we  have 
the  following  expressions; 


For  the  angles. 
Cos  A=R  X 


Cos  B=R  X 


Cos  C=Rx 


26c 
2ac 
2^6 


For  the  sides. 

26c  Cos  A\ 

a=v/(62-f-c2 -^ \ 

2ac  Cos  B 
6  =  v/  a^-fc-~ 


Uos  ti\ 

-rT-) 


c=v/(a2-|-62 


2a6  Cos  C 


uosu\ 

R— ; 


These  formulae  are  useful,  in  many  trigonometrical  investi- 
gations; but  are  not  well  adapted  to  logarithmic  computa- 
tion. 

Case  IV. 

When  the  three  sides  of  a  triangle  are  given,  the  angles 
may  be  found,  by  either  of  the  following  theorems;  in  which 
a.  S,  and  c  are  the  sides,  A,  B,  and  C,  the  opposite  angles, 
and  A=halfthe  sum  of  the  sides. 


2R 


J 


Sin  A=  -T-x/h{h-a)(h-b)(h-c) 


be 
2R 


Theorem  C.   i  Sin  B= ^Z  h{h~-a){h-b){h-c 

2R 


[Sm  C=  -^  x/  h{h-a){h-b){h-c) 


The  quantities  under  the  radical  sign  are  the  same  in  all 
the  equations. 

In  the  triangle  ACD,  (Fig.  23.) 
R  :  i::Sin  A  :p.     Therefore,  Sin  Ax6=Rxp. 

^^^     _^^b^c^-{b^^+c'~^^\  (Art.  221.  p.  105.) 

•'^  2c 


OF  TRIANGLES. 


1'>r'^ 


This,  by  the  reductions  in  page  106,  becomes 

V2hx2{h  -  a)X2{h-b)x2{h-c) 
P= 2c ^ 

Substituting  this  value  of/?,  and  reducing, 

2R 

Sin  A=-7 — \/h{h  —  a){h  —  b)(h  —  c) 

The  arithmetical  calculations  may  be  made,  by  adding  the 
logarithms  of  the  factors  under  the  radical  sign,  dividing  the 
sura  by  2,  and  to  the  quotient,  adding  the  losiarithms  of  radi- 
us and  2,  and  the  arithmetical  complements  of  the  logarithms 
of  6  andc.  (Arts.  39,  47,  59.) 

Ex.  Given  a  =  l34, 6  =  108,  and  c=80,  to  find  A,  B,  and  C. 
For  the  angle  A. 

A  161  log.  2.2068259 

/i-a  27  log.  1.4313638 

h-b  53  log.  1.7242759 

fi-c  81  lo-.  1.9084850 


RX2 


2)7.2709506 

*  3.6354753 

loz.  10.3010300 


108     a. 
80    a. 


13.9365053 
c.  7.9665762 
c.  8.0969100 


For  the 

a  134         a 
c     80         a 

angle  B. 

13.9365053 

c.  7.8728952 

.c.  8.0969100 

SiaB. 

9.9063105 

B^53°42'9" 

For  the  angle  C- 

13.9365053 
a  134         a.  c.  7.87289.52 
b  108         a.  c.  7.9665762 

Sin  A.  9.9999915 


Sin  C.  9.7759767 

0=36°  39' 20" 


Si„jA=Rv/^-^*)^ 


Theorem  D.   <(  Sin  iB=  Rv/ 


Sin  iC=  Rv/ 


{h-a){h-c) 
ac 

ah 


By  art.  210,  2  Sin2iA=R»  -Rxcos  A. 
Substituting  for  cos  A,  its  value,  as  given  in  page  132, 

624-c2-a2 


2SinHA=R2-R2X 


26c 


*  This  is  the  logarithm  of  the  area  of  the  triangle.  (Art.  222.^ 


134  PARTICULAR   SOLUTIONS 

Therefore  2Sin2iA=R2  x ^^ 

Bui2bc-\'a^-b''-c^=a^  -(h-cy={a+h-c){a-h-{-c) 
(Alg.  2o5.) 

Putting  then  A=|(a+6-|-c),  reducing,  and  extracting, 

Sin  i  A=Rv/ Y^ 

Ex.   Given  a,  5,  and  r,  as  before,  to  find  A  and  B. 
For  the  angle  A.  For  the  angle  B. 

h-b     53             }.l2iTio9  h-a    27  1.4313638 

h—c     81             1.^)084850  h-c    81  1.9084850 

b         108  a.  c.  *;.:•.«  »762  a         134  a.  c.  7,872,8952 

c           80  a.  c.  8.<:.-r9!00  c           80  a.  c.  8.0969100 


2)19.(;962471  2)19.8096540 

SiniA  9.8481235  Sin^B  9  654H270 


A=::89''38' 31"  B=53''42'9'' 

[cosiA=R/-(^^ 

TH.OH.M    E.     i  (,^^  .B^R^^CLJO 

icosiC=Rv/'^^ 

By  art.  210,  2Cos^iA=R2+Rxcos  A, 

Substituting  and  reducing,  as  in  the  demonstration  of  the 
last  theorem, 

25c4-&-+c2-«2              (b-^-c+aYb-^c-a) 
2Cos-A=R^  X -,,-^ =R^X^- i- -^ 


Pulling  h=l{a-{-b-\-c)  reducing  and  extracting, 

h(h-c) 

Cos  jA=Rv/  ^   , 

*  be 

Ex.  Given  the  sides  134,  108,  80;  to  find  B  and  C. 


OF  TRIANGLES. 


l3;-» 


For  the  angle  B. 

k  161  2.2068259 

h-b  53  1.7242759 

a  134  a.  c.  7.8728952 

c  80  a.  c.  8.0969100 


2)19.9009070 
Cos  iB  9.9504535 

B==53*42'9" 


For  the  angle  C. 

h  161  2.2068259 

A— c  81  1.9084850 

a  134  a.  c.  7  8728952 

b  108  a.  c.  7.9665762 


2)19.9547823 
CosiC  9.9773911 

0=36"  39' 20" 


TuEO 


REM 


F. 


I  Tan  .A=R^-^^^ 


\  Tan  iB=Rv/ 

I 
lTan^C=Rv/ 


{h-a){h-c) 

h{h-b) 
(^k-a){h-b) 


h(h-c) 


The  tangent  is  equal  to  the  product  of  radius  and  the  sine, 
divided  by  the  cosines.  (Art.  216.)  By  the  last  two  theorems, 
then, 


Ta„xA=^''"^^ 


COS 


RV 


(h-b){h-c) 


be 


~Rv/ 


h(h  —  a) 


be 


Thatis,Tan^A=Rv^ 


(h-bXh-c) 


h{h-a) 

Ex.  Given  the  sides  as  before,  to  find  A  and  C 
For  the  angle  A. 

k-b  53            1.7242759 

k-c  81              1.9084850 

h-a  27  a.  c.  8.5686-362 

h  161  a.  c.  7.7931741 


For  the  angle  B. 

h—a  27  1.4313638 

h-b  53  17242759 

h—c  81  a.  c.  8.0915150 

k  161  B.C.  7.7931741 


Tan  iA 


2)19.9945712 
9.9972856 


A=89°38'3r 


Tan  iC 

C=36°39'  20" 


2)19.0403288 
9.5201644 


The  three  last  theorems  give  the  angle  required,  tvithout 
ambiguity.     For  the  half  of  any  angle  must  be  less  than  90.° 

Of  these  different  methods  of  solution,  each  has  its  advan- 
tages in  particular  cases.  It  is  expedient  to  find  an  angle, 
sometimes  by  its  sine,  sometimes  by  its  cosine,  and  some- 
times by  its  tangent. 

By  the  first  of  the  four  preceding  theorems  marked  C,  D, 
E,  and  F,  the  calculation  is  made  for  the  sine  of  the  whole  an- 
gle ;  by  the  others,  for  the  sine,  eosine,  or  tangent,  of  hnjfihf' 


136  PARTICULAR  SOLUTIONS,  &c. 

angle.  For  finding  an  angle  near  90°,  each  of  the  three  last 
theorems  is  preferable  to  the  first.  In  the  example  above,  A 
would  have  been  uncertain  to  several  seconds,  by  theorem  C, 
if  the  other  two  angles  had  not  been  determined  also. 

But  for  a  very  small  angle,  the  first  method  has  an  advan- 
tage over  the  others.  The  third,  by  which  the  calculation  is 
made  for  the  cosine  of  half  the  required  angle,  is  in  this  case 
the  most  defective  of  the  four.  The  second  will  not  answer 
well  for  an  angle  which  is  almost  180°.  For  the  Aa//*of  this 
is  almost  90°  ;  and  near  90°,  the  differences  of  the  sines  are 
very  small. 


notes- 


Note  A.  Page  1 


rpHE  7iame  Loo:arithm  is  from  >^oyog  ratio,  and  a^i&ixos  num- 
ber. Considering  the  ratio  of  a  to  1  as  a  simple  ratio, 
that  of  a^  to  1  is  a  duplicate  ratio,  of  a^  to.l  a  triplicate  ra- 
tio, &:c.  (A!g.  354.)  Here  the  exponents  or  logarithms  2,  3, 
4,  &ic,  show  how  many  times  the  simple  ratio  is  repeated  as  a 
factor,  to  form  the  compound  ratio.  Thus  the  ratio  of  100 
to  1,  is  the  square  of  the  ratio  of  10  to  1  ;  the  ratio  of  1000 
to  1 ,  is  the  cube  of  the  ratio  of  1 0  to  1 ,  &tc.  On  this  account, 
logarithms  are  called  the  iiieasurts  of  ratios  ;  that  is  of  the 
ratios  which  different  numbers  bear  to  unity.  See  the  In- 
troduction to  Hutton's  Tables,  and  Mercator*s  Logarithmo- 
Technia,  in  Masseres'  Scriptores  Logarithmici. 


Note  B.  p.  4. 

If  1  be  added  to  -.09691,  it  becomes  1 —-09691,  which  is 
equal  to  -f-'90309.  The  decimal  is  here  rendered  positive, 
by  subtracting  the  figures  from  1.  But  it  is  made  1  too  great. 
This  is  compensated,  by  adding  -  1  to  the  intes:ral  part  oY 
the  logarithm.     So  that  — 2— .09691  =  — 3+.90309. 

In  the  same  manner,  the  decimal  part  of  any  logarithm 
which  is  wholly  negative,  may  be  rendered  positive,  by  sub- 
tracting it  from  l,and  adding  —1  to  the  index.  The  sub- 
traction is  most  easily  performed,  by  taking  the  right  hand 
significant  figure  from  10,  and  each  of  the  other  figures  from 
9.      (Art.  55.) 

On  the  other  hand,  if  the  index  of  a  logarithm  be  negative, 
while  the  decimal  part  is  positive  ;  the  whole  may  be  render- 
ed negative,  by  subtracting  the  decimal  part  from  1,  and  ta- 
king —  1  from  the  index. 


to 


138  TRIGONOMETRY 


Note  C.  p.  7. 


It  is  common  to  define  logarithms  to  be  a  series  of  num- 
bers in  arithmetical  progression,  corresponding  with  another 
series  in  geometrical  progression.  This  is  calculated  to  per- 
plex the  learner,  when,  upon  opening  the  tables,  he  finds  that 
the  natural  numbers,  as  thej  stand  there,  instead  of  being  in 
geometrical^  are  in  arithmetical  progression  ;  and  that  the  log- 
arithms are  not  in  arithmetical  progression. 

It  is  true,  that  a  geometrical  series  may  be  obtained,  by 
taking  out,  here  and  there,  a  few  of  the  natural  numbers  ; 
and  that  the  logarithms  of  these  will  form  an  arithmetical  se- 
ries. But  the  definition  is  not  applicable  to  the  whole  of  the 
numbers  and  logarithms,  as  they  stand  in  the  tables. 

The  supposition  that  positive  and  negative  numbers  have 
the  same  series  of  logarithms,  (p.  7.)  is  attended  with  some 
theoretical  difficulties.  But  these  do  not  affect  the  practical 
rules  for  calculating  by  logarithms. 


Note  D.  p.  43. 

To  revert  a  series,  of  the  form 

X  =:^un-\-bn^  -\-cn^  ■\-dn'^  -\-en^  -]- &c. 

that  is,  to  find  the  value  of??,  in  terms  of  ac,  assume  a  series, 
with  indeterminate  co-efficients,  (Alg.  490.  6.) 

Letn=Aa:  +  Bx2-fCa:34-Dx^-i-Ex^-f-&c. 

Finding  the  powers  of  this  value  of  ?i,  by  multiplying  the 
series  into  itself,  and  arranging  the  several  terms  according 
to  the  powers  of  x  ;  we  have 

n^=A2a:2  4-2AB.x^+2AC>       -I-2BC  >     5  .&. 
^3=  A^r34-3A-Bx^    +3A*C>     ,  ,  « 

J    >  X     -f-KC. 


n'  = 


+  3AB' 

A^r*    -f 4A»Ba;5-|-&c. 

A^T^-f&C. 


NOTES. 


139 


Substiluting  these  values,  for  ?i  and  its  powers,  in  the  first 
series  above,  we  have 


„l 


faA  x+aB   }     ,-\-aC       )       +        «D^       +     aE 
I  +iA2  5  ^   +26AB>  x^+  '2b AC  |       +26BC 

+  cAM       +     ^B^   )>x'-\-2bAD 


-f.BcA^B  I       -\-3cA'C  [x' 
-f-     dA' J       -{-3cAB^ 
-f4JA3B 

-h      eA^  J 


Transposing  x,  and  making  the  co-efficients  of  the  several 
powers  of  x  each  equal  to  0,  we  have 

aA  -  1  =0, 

«B+6A^=0, 

uC-\-nAB-\-cA^=0, 

aD4-26AC  +  6B2-h3cA2B  +  rfA*=0, 

a.E-\-2bBC-\'2bAD-\-3cA^C  +  3cAB^  -\-^dA^B-^eA'  =:0. 


And  reducing  the  equations, 


A  =  i 
B=- 


C= 


D=-- 


26^  —ac 


5b^—5abc+a''d 


E=b 


l^b'—^^ab^'c-^'Sa^c^+ea^bd—a^e 


These  are  the  values  of  the  co-efficients  A,  B,  C,  &c.  in 
ihe  assumed  series 

n=Ax-fBx2+Cx   +Dx^-|-Ex5+&c. 

Applying  these   results  to   the   logarithmic  series.     (Art. 
€C.  p.  43.) 


140  TRIGONOMETRY- 

in  which 

we  have,  in  the  inverted  series 

A=^=l  D=— L- 


2.3.4 
B=  -6  =  1 


2.3  2.3.4.5 

Therefore 

2  ^2.3  ^2.3.4  ^2.3.4.5  ^^^* 


Note  E.  p.  50. 

According  to  the  scheme  lately  introduced  into  France,  of 
dividing  the  denominations  of  weights,  measures,  &c.  into 
tenths,  hundredths,  &;c.  the  fourth  part  of  a  circle  is  divided 
into  100  degrees,  a  degree  into  100  minutes,  a  minute  into 
100  seconds,  &c.  The  whole  circle  contains  400  of  these 
degrees  ;  a  plane  triangle  200.  If  a  right  angle  be  taken 
for  the  measuring  unit  ;  degrees,  minutes,  and  seconds,  may 
be  written  as  decimal  fractions.  Thus  36°  5'  49''  is  0.360549. 

C      10°=9°      ^ 
According  to  the  French  division    <    100' =54'     >  English. 

/  1000 '  =  324"  S 


Note  F.  p.  82. 

If  the  perpendicular  be  drawn  from  the  angle  opposite  the 
longest  side,  it  will  always  fall  within  the  triangle  ;  because 
the  05 her  two  angles  must,  of  course,  be  acute.  But  if  one 
of  the  angles  at  the  base  be  obtuse^  the  perpendicular  will  fall 
without  the  triangle,  as  CP,  (Fig.  38.) 

In  this  case,  the  side  on  which  the  perpendicular  falls,  is 
to  the  sum  of  the  other  two  ;  as  the  difference  of  the  latter. 
to  i\iQ  Sim  of  the  segments  made  by  the  perpendicular. 


NOTES.  141 

The  demonstration  is  the  same,  as  in  the  other  case,  ex- 
cept that  AH  =  BP+PA,  instead  of  BP-PA. 

Thus  in  the  circle  BDHL(Fig.  38.)  of  which  C  is  the  cen- 
tre, 

ABxAH=ALxAD;  therefore  AB  :  AD::AL  :  AH. 

But  AD  =  CD-f-CA=CB+CA 
And  AL=CL-CA=CB-CA 
And  AH=HP4-PA  =  RP+PA 

Therefore 
AB  :CB-}-CA::CB-CA  :  BP+PA. 

When  the  three  sides  are  given,  it  may  be  known  whether 
one  of  the  angles  is  obtuse.  For  any  angle  of  a  triangle  is 
obtuse  or  acute,  according  as  the  square  of  the  sine  subtend- 
ing the  angle  is  greater,  or  less,  thsin  the  sum  of  the  squares  of 
the  sides  containing  the  angle.  (Euc.  12,  13.  2.) 


Note  G.  p.   104. 

Gunter's  Sliding  Rule  is  constructed  upon  the  same  prin- 
ciple as  his  scale,  with  the  addition  of  a  slider,  which  is  so 
contrived  as  to  answer  the  purpose  of  a  pair  of  compasses,  in 
working  proportions,  multiplying,  dividing,  &z;c.  The  lines  on 
the  fixed  part  are  the  same  as  on  the  scale.  The  slider  con- 
tains two  lines  of  numbers,  a  line  of  logarithmic  sines,  and  a 
line  of  logarithmic  tangents. 

To  multiply  by  this,  bring  1  on  the  slider,  against  one  of 
the  factors  on  the  fixed  part ;  and  against  the  other  factor  on 
the  slider,  will  be  the  product  on  the  fixed  part.  To  divide, 
bring  the  divisor  on  the  slider,  against  the  dividend  on  the 
fixed  part  ;  and  against  1  on  the  slider,  will  be  the  quotient 
on  the  fixed  part.  To  work  ?l proportion,  bring  the  first  term 
on  the  slider,  against  one  of  the  middle  terms  on  the  fixed 
part ;  and  against  the  other  middle  term  on  the  slider,  will  be 
the  fourth  term  on  the  fixed  part.  Or  the  first  term  may  be 
taken  on  the  fixed  part  ;  and  then  the  fourth  term  will  be 
found  on  the  slider. 

Another  instrument  frequently  used  in  trigonometrical  con- 
ftructionsj  i< 


142  TRIGONOMETRY. 

The  Sector. 

This  consists  of  two  equal  scales,  moveable  about  a  point 
as  a  centre.  The  lines  which  are  drawn  on  it  are  of  two 
kinds  ;  some  being  parallel  to  the  sides  of  the  instrument, 
and  others  diverging  from  the  central  point,  like  the  radii  of 
a  circle.  The  latter  are  called  the  double  lines,  as  each  is 
repeated  upon  the  two  scales.  The  single  lines  are  of  the 
same  nature,  and  have  the  same  use,  as  tiiose  which  are  put 
upon  the  common  scale  ;  as  the  lines  of  equal  parts,  of  chords, 
of  latitude,  (Sic.  on  one  face  ;  and  the  logarithmic  lines  of 
numbers,  of  sines,  and  of  tangents,  on  the  other. 

The  double  lines  are 

A  line  o(  Lines,  or  equal  parts,  marked     Lin.  or  L, 

A  line  of  Chords,  Cho.  or  C. 

A  line  of  natural  Smc5,  Sin.  or  S. 

A  line  of  natural  Tangents  to  45^.  Tan.  or  T. 

A  line  of  tangents  above  45°.  Tan.  or  T. 

A  line  of  natural  Secants^  Sec.  or  S. 

A  line  o( Polygons,  Pol.  or  P. 

The  double  lines  o( Chords,  o( sines,  and  oi tangents  to  45°, 
are  all  of  the  same  radius  ;  beginning  at  the  central  point, 
and  terminating  near  the  other  extremity  of  each  scale  ; 
the  chords  at  G0%  the  sines  at  90°,  and  the  tangents  at  45°. 
(See  art.  95.)  The  line  of  lines  is  also  of  the  same  length, 
containing  ten  equal  parts  which  are  numbered,  and  which 
are  again  subdivided.  The  radius  of  the  lines  of  secants, 
and  of  tangents  above  45°,  is  about  one  fourth  of  the  length 
of  the  other  lines.  From  the  end  of  the  radius,  which  for 
the  secants  is  at  0,  and  for  the  tangents  at  45°,  these  lines 
extend  to  between  70°  and  80°.  The  line  of  polygons  is 
numbered  4,  5,  6,  &;c.  from  the  extremity  of  each  scale,  to- 
wards  the  centre. 

The  simple  principle  on  which  the  utility  of  fhese  several 
pairs  of  lines  depends  is  this,  that  the  sides  of  similar  triangles 
are  proportional.  (Euc.  4.  6.)  So  that  sines,  tangents,  &;c. 
are  furnished  to  any  radius^  within  the  extent  of  the  opening 
of  the  two  scales.  Let  AC  and  AC  (Fig.  40.)  be  any  pair  of 
lines  on  the  sector,  and  AB  and  AB'  equal  portions  of  these 
lines.  As  AC  and  AC  are  equal,  the  triangle  ACC  is  isos- 
ceks,  and  similar  to  ABB'.     Therefore, 

AB:  AC::BB':CC'. 


NOTES.  143 

Distances  measured  from  ihe  centre  on  either  scale,  as 
AB  and  AC,  are  called  lateral  distances.  And  the  distances 
between  corresponding  points  of  the  two  scales,  as  BB'  and 
CC  are  called  transverse  distances. 

Let  AC  and  CC  be  radii  of  two  circles.  Then  if  AB  be 
the  chord,  sine,  tangent,  or  secant,  of  any  number  of  degreei? 
in  one  ;  BB'  will  be  the  chord,  sine,  tangent,  or  secant,  of 
the  same  number  of  degrees  in  the  other.  (Art.  119.)  Thus, 
to  find  the  chord  of  30°,  to  a  radius  of  four  inches,  open  the 
sector  so  as  to  make  the  transverse  distance  from  60  to  60, 
on  the  lines  of  chords,  four  inches  ;  and  the  distance  from 
30  to  30,  on  the  same  lines,  will  be  the  chord  required.  To 
find  the  sine  of  23°,  make  the  distance  from  90  to  90,  on  the 
lines  of  sines,  equal  to  radius  ;  and  the  distance  from  28  to 
28  will  be  the  sine.  To  find  the  tangent  of  37"^,  make  the  dis- 
tance from  45  to  45,  on  the  lines  of  tangents,  equal  to  radius  ; 
and  the  distance  from  37  to  37  will  be  the  tangent.  In  find- 
ing secants,  the  distance  from  0  to  0  mu?t  be  made  radius. 
(Art.  201.) 

To  lay  down  an  angle  of  34°,  describe  a  circle,  of  any  con- 
venient radius,  open  the  sector,  so  that  the  distance  from  GO 
to  GO  on  the  lines  of  chords  shall  be  equal  to  this  radius,  and 
to  the  circle  apply  a  chord  equal  to  the  distance  from  34  to 
34.  (Art.  161.)  For  an  angle  above  60°,  the  chord  of  Ac//' 
the  number  of  degrees  may  be  taken,  and  applied  twice  on 
the  arc,  as  in  art.  161. 

The  line  o{ polygons  contains  the  chords  of  arcs  of  a  circle 
which  is  divided  into  equal  portions.  Thus  the  distances 
from  the  centre  of  the  sector  to  4,  5,  6,  and  7,  are  th<i  chords 
^^  }•>  ji  h  ^"^  4  '^^  ^  circle,  The  distance  6  is  the  radius. 
(Art.  95.)  This  line  is  used  to  make  a  regular  polygon,  or  to 
inscribe  one  in  a  given  circle.  Thus,  to  make  a  pentagon 
with  the  transverse  distance  from  6  to  G  for  radius,  describe  a 
circle,  and  the  distance  from  5  to  5  will  ha  the  length  of  one 
of  the  sides  of  a  pentagon  inscribed  in  that  circle. 

The  line  of  lines  is  used  to  divide  a  line  into  equal  or  pro- 
portional parts,  to  find  fourth  proportionals,  &,c.  Thus,  t© 
divide  a  line  into  7  equal  parts,  make  the  length  of  the  given 
line  the  transverse  distance  from  7  to  7,  and  the  distance 
from  1  to  1  will  be  one  of  the  parts.  To  find  f  of  a  line, 
make  the  transverse  distance  from  5  to  5  equal  to  the  given 
line  ;  and  the  distance  from  3  to  3  will  be  f  of  it. 

In  working  the  proportions  in  trigonometry  on  the  sector, 
the  lengths  of  the  sides  of  triangles  are  tfiken   from  the   linf 


144  TRIGONOMETRY. 

of  lines,  and  the  degrees  and  minutes  from  the  lines  of  sines, 
tangents,  or  secants.     Thus  in  art.  135j  ex.  1, 

35  :  R::26  :  Sin  48°. 

To  find  the  fourth  term  of  this  proportion  by  the  sector, 
make  the  lateral  distance  35  on  the  line  of  lines,  a  transverse 
distance  from  90  to  90  on  the  lines  of  sines  ;  then  the  lateral 
distance  26  on  the  line  of  lines,  will  be  the  transverse  dis- 
tance from  48  to  48  on  the  lines  of  sines. 

For  a  more  particular  account  of  the  construction  and  uses 
of  the  Sector,  see  Stone's  edition  of  Bion  on  Mathematical 
Instruments,  Button's  Dictionary,  and  Robertson's  Treatise 
on  Mathematical  Instruments. 


Note  H.  p.  124. 

The  errour  in  supposing  that  arcs  less  than  1  minute  are 
proportional  to  their  sines,  can  not  affect  the  tirst  ten  places 
of  decimals.  Let  AB  and  AB'  (Fig.  41.)  each  equal  1  min- 
ute. The  tangents  of  these  arcs  BT  and  B'T  are  equal,  as 
are  also  the  sines  BS  and  B'S.  The  arc  BAB'  is  greater 
than  BS  +  B'S,  but  less  than  BT  +  BT.  Therefore  BA  is 
greater  than  BS,  but  less  than  BT  :  that  is,  the  difference  be- 
tween the  sine  and  the  arc  is  less  than  the  difference  between  the 
sine  and  the  tangent, 

Now  the  sine  of  1  minute  is        0.000290888216 
And  the  tangent  of  1  minute  is    0.000290888204 

The  difference  is  0.00000000001 2 


The  difference  between  the  sine  and  the  arc  of  1  minute 
is  less  than  this  •,  and  the  errour  in  supposing  that  the  sines 
of  1',  and  of  0'  52''  44'"  3''"  45"'"  are  proportional  to  their 
arcs,  as  in  art.  223,  is  still  less. 


Note  I.  p.   125. 

There  are  various  ways  in  which  sinas  and  cosines  may  be 
more  expeditiously  calculated,  than   by   the  method  which  is 


NOTES.  146 

given  here.  But  as  we  are  already  supplied  with  accurate 
trigonometrical  tables,  the  computation  of  the  canon  is,  to 
the  great  body  of  our  students,  a  subject  of  speculation,  rath- 
er than  of  practical  utility.  Those  who  wish  to  enter  into  a 
minute  examination  of  it,  will  of  course  consult  the  treatises 
in  which  it  is  particularly  considered. 

There  are  also  numerous  formulas  of  veriftcatioji,  which 
are  used  to  detect  the  errours  with  which  any  part  of  the  cal- 
culation is  liable  to  be  affected.  For  these,  see  Legendre's 
and  Woodhouse's  Trigonometry,  Lacroix's  Differential  Cal- 
culus, and  particularly  Euler's  Analysis  of  infinites. 

Note  K.  p.  127. 

The  following  rules  for  finding  the  sine  or  tangent  of  a  very 
small  arc,  and,  on  the  other  hand,  for  finding  the  arc  from  its 
sine  or  tangent,  are  taken  from  Dr.  IViaskelyne's  Introduction 
to  Taylor's  Logarithms. 

To  find  the  logarithmic  sine  of  a  very  small  arc. 

From  the  sum  of  the  constant  quantity  4.6855749,  and  the 
logarithm  of  the  given  arc  reduced  to  seconds  and  decimals, 
subtract  one  third  of  the  arithmetical  complement  of  the  log- 
arithmic cosine. 

To  find  the  logarithmic  tangent  of  a  very  small  arc. 

To  the  sum  of  the  constant  quantity  4.6855749,  and  the 
logarithm  of  the  given  arc  reduced  to  seconds  and  decimals, 
add  two  thirds  of  the  arithmetical  complement  of  the  logarith- 
mic cosine. 

To  find  a  small  arc  from  its  logarithmic  sine. 

To  the  sum  of  the  constant  quantity  5.3144251,  and  the 
given  logarithmic  sine,  add  one  third  of  the  arithmetical  com- 
plement of  the  logarithmic  cosine.  The  remainder  diminish- 
ed by  10,  will  be  the  logarithm  of  the  number  of  seconds  in 
the  arc. 

To  find  a  small  arc  from  its  logarithmic  tangent. 

From  the  sum  of  the  constant  quantity  5.3144251,  and  the 
given  logarithmic  tangent,  subtract  two  thirds  of  the  arithmet- 
ical complement  of  the  logarithmic  cosine.  The  remainder 
diminished  by  10,  will  be  the  logarithm  of  the  number  of  see- 
onds  in  the  arc. 

For  the  demonstration  of  these  rules,  see  Woodhouse's 
Trigonometry,  p.  189. 

20 


A  TABLE  OF 


NATURAL  SINES  AND  TANGENTS; 


TO  EVERY  TEN  MINUTES  OF  A  DEGREE, 


IF  the  given  angle  is  less  than  45°,  look  for  the  title 
of  the  column,  at  the  top  of  the  page  ;  and  for  the  de- 
grees and  minutes,  on  the  left.  But  if  the  angle  is  be- 
tween 45°  and  90°,  look  for  the  title  of  the  column,  at 
the  bottom  ;  and  for  the  degrees  and  minutes,  on  the 
right. 


148 


NATURAL  SINES  AND  TANGENTS 


D.  M. 


0°  0' 
10 
20 
30 
40 

0°  50' 


Sine 


O.OOOOOOO 
0029089 
0058177 
00S7265 
0116353 
0145439 


r 

0' 

0.0174524 

10 

0203608 

20 

0232690 

30 

0^01769 

40 

0290847 

r 

50' 

0319922 

go 

0' 

0.0348995 

10 

0378065 

20 

0407131 

30 

•  0436194 

40 

0465253 

QO 

50' 

0494308 

'^° 

0' 

0.0523360 

10 

0552406 

20 

0581448 

30 

0610485 

40 

0639517 

3° 

50' 

0668544 

40 

0' 

00697565 

10 

0726580 

20 

0755589 

30 

0784591 

40 

0813587 

4° 

50' 

0842576 

5° 

0' 

0.0871557 

10 

0900532 

20 

0929499 

30 

0958458 

40 

0987408 

5° 

50' 

10'6351 

D. 

Cosine 

rHn<!;ent    1  Cotfin2;ent 


O.OOOOOUO 
>  0029089 
0058178 
0087269 
0116361 
0146464 

0.0174551 
0203650 
0232753 
0261859 
0290970 
0320^86 

0.0349208 
0378335 
0407469 
0436609 
0465757 
0494913 

0.0524078 
0553251 
0582434 
0611626 
0640829 
0670043 

0.0699268 
0728605 
0757755 
0787017 
0816293 
0848583 

0.0874887 
0904206 
0933340 
0962890 
0992257 
1021641 

(.'otansent 


Infinite 
343.77371 
171.88540 
114.58865 
85.939791 
68.750087 

57.289962 
49.103881 
42.964077 
38.188459 
34.367771 
31.241577 

28.636253 
26.431600 
24.541758 
22.903766 
21.470401 
20.205553 

19.081137 
18.074977 
17.169337 
16.349855 
15  604784 
14.924417 

14.300666 
13.726738 
13.196883 
12.706205 
12.250505 
11.826167 

11.430052 
11.059431 
10.711913 
16.385397 
10.078031 
9.7881732 


Cosii 


Tangent 


1.0000000 
0.9999958 
9999831 
9999619 
9999323 
9998942 

0.9998477 
9997927 
9997292 
9996573 
9995770 
9994881 

0.9993908 
9992851 
9991*709 
9990482 
9989171 
9987775 

0.9986295 
9984731 
9983082 
9981348 
9979630 
9977627 

0.9975641 
9973669 
9971413 
9969173 
9966849 
9964440 


D.  M. 


90°  0' 
50 
40 
30 
20 

89°  10 

89°  0 
60 
40 
30 
20 

88°  10' 

88°  0' 
50 
40 
30 
20 

87°  10' 

87°  0' 
50 
40 
30 
20 

86°  10' 

86°  0' 
50 
40 
30 
20 

85°  10' 


0.9961947 

85° 

0 

9950370 

50 

9956708 

40 

9953962 

30 

9951132 

20 

9948217 

84° 

10 

Sine 

D. 

M. 

NATURAL  SINES  AND  TANGENTS. 


149 


;  D.  51. 

10 

20 

30 

40 

i  6o  50' 

I  7°  0 
i    10 

20 
,    30 

40 
I  7°  5o' 

i  8°  0' 

!   10 

20 

30 

40 

j  8°  50' 

f  9°  0' 
10 
20 
30 

I  40 
9°  50' 


10' 


10= 


11 


0' 
10 
20 
30 
40 
50' 


Sine 


0' 
10 
20 
30 
40 
11°  50' 


D.  M. 


0.1045285 
1074210 
1103126 
1132032 
1160929 
1189816 

0.1218693 
1247560 
1276416 
1305262 
1334096 
1362919 

0.1391731 
1420531 
14493 19 
1478094 
1506857 
1535667 

0.1664345 
1593069 
1621779 
1650476 
1679159 
1707828 

0.1736482 
1765121 
1793746 
1822355 
1850949 
1879528 

O.J  908090 
1936636 
1965166 
1993679 
2022176 
2050655 


Cosine 


Tangent. 


0.1051042 
1080462 
1109899 
1139356 
1168832 
1198329 

0.1227846 
1257384 
1286943 
1316525 
1346129 
1375757 

0.1404085 
1435084 
1464784 
1494510 
1524262 
1554040 

0.1583844 
1613677 
1643537 


Cotangent 


9  5143645 
9.2553035 
9.0098261 
8.7768874 
8.5555468 
8.3449558 

8.1443464 
7.9530224 
7.7703506 
7.5957541 
7.4287064 
7.2687255 

7.1153697 
6.9682335 
6.8269437 
6.6911562 
6.5605538 
6.4348428 

6.3137515 

6-1970279 
6.0844381 


1703344  I  5.8708042 
1733292  5.7693688 


0.1763270 
1793279 
1823319 
1853390 
1883495 
1913632 

0.1943803 
1974008 
2004248 
2034523 
2064834 
2095181 


5.6712818 
5.5763786 
5.4845052 
5.3955172 
5.3092793 
5.2256647 

5.1445540 
5.0658352 
4.9894027 
4.9151570 
4.8430045 
4.7728568 
Cotangent  |  Tangent 


Cosine 

D. 

84° 

M. 

0' 

0.9945219 

9.942136 

50 

9938969 

40 

9935719 

30 

9932384 

20 

9928065 

830  10' 

0.9925462 

83° 

0' 

9921874 

50 

9918204 

40 

9914449 

» 

30 

9910610 

20 

9906687 

82° 

10' 

0.9902681 

82° 

0' 

9898590 

50 

9894416 

40 

9890159 

30 

9885817 

20 

9881392 

81° 

10' 

0.9876883 

81° 

0' 

9872291 

50 

9867615 

40 

9862856 

30 

9858013 

20 

9853087 

80° 

10' 

0.9848078 

80° 

0' 

9842985 

50 

9837808 

40 

9832549 

30 

9827206 

20 

9821781 

79° 

10' 

0.9816272 

79° 

0' 

9810680 

50 

9805005 

40 

9799247 

30 

9793406 

20 

9787483 

78° 

10' 

Sine   «  D.  M. 


150 


NATURAL  SINES  AND  TANGENTS. 


D.  M. 

12°  0~ 
10 
20 
30 
40 

12°  60' 


13°  O 
10 
20 
30 
40 

13°  50' 

14°  0' 

10 
20 
30 
40 
14°  50' 

15°  0 
10 
20 
30 
40 
60' 

0' 
10 
20 
30 
40 
50' 

0' 

10 
20 
30 
40 
60' 


Sine 


16= 


16= 


17' 


-17' 


D.  M. 


0.2079117 
2107561 
2135988 
2164396 
2192786 
2221168 

0.2249511 
2277844 
2306159 
2334454 
2362729 
2390984 

0.2419219 

2447433 
2475627 
2503800 
2531952 
2560082 

0.2588190 
2616277 
2644342 
2672384 
2700403 
2728400 

0.2756374 

2784324 

2812251 

2840153 

2868032! 

2895887 


Tangent 


0.2923717 
2951522 
2979303 
3007058 
3034788 
3062492 


Cosine 


0.2125566 
2165988 
2186448 
2216947 
2^247485 
2278063 

0.2308682 
2339342 
2370044 
2400788 
2431575 
2462405 

0.2493280 
2524200 
2555165 
2586176 
2617234 
2648339 

0.2679492 
2710694 
2741945 
2773245 
2804597 
2835999 

0.2867454 
2898961 
2930521 
2962135 
2993803 
3026527 

0.3057307 
30S9143 
3121036 
3152988 
3184998 
3217067 


Cotangent 


Cotnn?ent 


4.7046301 
4.6382457 
4.5736287 
4.5107085 
4.4494181 
4.3896940 

4.3314759 
4.2747066 
4.2193318 
4.1652998 
4.1125614 
4.0610700 

4.0107809 
3.9616518 
3.9136420 
3.8667131 
3.8208281 
3.7759519 

3.7320508 
3.6890927 
3.6470467 
3.6058835 
3.5655749 
3.5260938 

3.4874144 
3.4495120 
3.4123626 
3.3759434 
3.3402326 
3.3052091 

3.2708526 
3.2371438 
3.2040638 
3.1715948 
3.1397194 
3.1084210 


Cosine    1).  IVi 


09781476 
9775387 
9769215 
9762960 
9756623 
9750203 

0.9743701 
9737116 
9730449 
9723699 
9716867 
9709963 

0.9702957 
9695879 
9688719 
9681476 
9674152' 
9666746 

0.9659258 
9651689 
9644037 
9636305 
9628490 
9620594 

0.9612617 
9604558 
9596418 
95^8197 
9579895 
9571612 

0.9563048 
9554502 
9545876 
9537170 
9528382 
9519514 


18' 


IT 


0' 
50 
40 
30 
20 
10 


77°  0' 
60 
40 
30 
20 

76°  10' 

76°  0' 
50 
40 
30 
20 

75°  10' 


0 
50 
40 
30 
20 
10 


74= 


74°  O' 
50 
40 
30 
20 

75°  10' 


Tangent 


i^ine 


73' 


72' 


0' 
50 
40 
30 
20 
10' 


D.  M 


NATURAL  SINES  AND  TANGENTS. 


151 


D.  M.  1 

bine 

Tangent 

Cotangent 

Cosine 

72°  0' 

18°  0' 

0.3090170 

0.3249197 

3.0776835 

0.9510565 

10 

3117822 

3281387 

3.0474915 

9501536 

50 

20 

3145448 

3313639 

3.0178301 

9492426 

40 

30 

3173047 

3345953 

2.9886850 

94i>3237 

30 

40 

3200619 

3378330 

2.9600422 

9473966 

20 

18°  50' 

3228164 

3410771 

2.9318885 

9464616 

71°  10' 

19°  0' 

0.3255682 

0.3443276 

2.9042109 

0.9155186 

71°  0' 

10 

3283172 

3475846 

2.8769970 

9445675 

50 

20 

3310634 

3508483 

2.8502349 

9436085 

40 

30 

3338069 

3541186 

2.8239129 

9426415 

SO 

40 

3365475 

3573956 

2.7980198 

9416665 

20 

19°  50' 

3392852 

3606795 

2.7725448 

9406835 

70°  10' 

20°  0' 

0.3420201 

0.3639702 

2.7474774 

0.93969261 70°  oj 

10 

3447521 

3672680 

2.7228076 

9386938 

50 

20 

3474812 

3705728 

2.6985254 

9376869 

40 

30 

3502074 

3738847 

2.67462 i 5 

9366722 

30 

40 

3529306 

3772038 

2.6510867 

9356495 

20 

20°  50' 

3556508 

3805302 

2.6279121 

9346189 

69°  10' 

21°  0' 

0.3583679 

0.3838640 

2.6050891 

0.9335804 

69°  0' 

10 

3610821 

3872053 

2.5826094 

9325340 

50 

20 

3637932 

3905541 

2.5604649 

9314797 

40 

30 

3665012 

3939105 

2.5386479 

9304176 

30 

40 

3692061 

3972746 

2.5171507 

9293475 

20* 

21°  50 

3719079 

4006465 

2.4959661 

9282696 

68°  10' 

22°  0' 

0.3746066 

0.4040262 

2.4750869 

0.9271839 

68°  0' 

10 

3773021 

4074139 

2.4545061 

9260902 

50 

20 

3799944 

4108097 

2.4342172 

9249888 

40 

30 

3826834 

4142136 

2.4142136 

9238795 

30 

40 

3853693 

4176257 

2.3944889 

9227624 

20 

22°  50 

3880518 

4210460 

2.3750372 

9216375 

67°  10' 

23°  0' 

0.3907311 

0.4244748 

2  3558524 

0.9205049 

67°  0' 

10 

3934071 

4279121 

2.3369287 

9193644 

50 

20 

3960798 

4313579 

2.3182606 

9182161 

40 

30 

3987491 

4348124 

2.2998425 

9170601 

30 

40 

4014150 

4382756 

2.2816693 

9158963 

20 

23°  50 

4040776 

4417477 

2.2637357 

9147247 

66°   10/ 
D.  M. 

D.  M. 

Cosine 

Cotangent 

Tangent  j 

iSine 

152 


NATURAL  SINES  AND  TANGENTS. 


D. 

M. 

24° 

.  0' 

10 

20 

30 

40 

24° 

50' 

25° 

0' 

10 

20 

30 

40 

25° 

50' 

26° 

0' 

10 

20 

30 

40 

26° 

50' 

27° 

0' 

10 

20 

30 

40 

27° 

50' 

28° 

0' 

10 

20 

30 

40 

280 

50' 

29° 

0' 

10 

20 

30 

40 

29° 

50' 

D. 

M. 

Sine 


0.4067366 
4093923 
4120445 
4146932 
4173335 
4199801 

0.4226183 

4252528 
4278838 
4305111 
4331348 
4357548 

0.4383711 
4409838 
4435927 
4461978 
4487992 
4513967 

0.4539905 
4565805 
4591665 
4617486 
4643269 
4669012 

0.4694716 

4720380 
4740004 
4771588 
4797131 
4822634 

0.4848096 
4873517 
4898897 
4924236 
4949532 
4974787 


Cosine 


Tangent 


0.4452287 
4487187 
4522179 
4557263 
4592439 
4627710 

0.4663077 
4698539 
4734098 
4769755 
4805512 
4841368 

0.4877326 
4913386 
4949549 
4985816 
5022189 
5058668 

0.5095254 
5131950 
5168755 
5205671 
5242698 
5279839 

0.6317094 
5354465 
5391953 
6429557 
5467281 
5605126 

0.5543091 
5581179 
5619391 
5657728 
5696191 
5734783 


Cotangent 


Cotangent 


2.2460368 
2.2285676 
2.2113234 
2.1942997 
2.1774920 
2.1608958 

2.1445069 
2.1283213 
2.1123348 
2.0965436 
2.0809438 
2.0665318 

2.0503038 
2.0352565 
2.0203862 
2.0056897 
1.9911637 
1.9768050 

1.9626106 
1.9485772 
1.9347020 
1.920982; 
1.9074147 
1.8939971 

1.8807266 
1  8676003 
1.8546159 
1.8417709 
1.8290628 
1.8164892 

1.8040478 
1.7917362 
1.7795524 
1.7674940 
1.7555690 
1.7437463 


Tangent 


Cosine 

D. 

M. 

0.9135455 

66' 

0' 

9123584 

50 

9111637 

40 

9099613 

30 

9087511 

20 

9075333 

65o 

10' 

0.9063078 

65° 

0' 

9050746 

50 

9038338 

40 

9025853 

30 

9013292 

20 

9000654 

64° 

10' 

0.8987940 

64° 

0' 

8975151 

50 

8962285 

40 

8949344 

30 

8936326 

20 

8923234 

63° 

10' 

0.8910065 

6.3° 

0' 

8896822 

50 

8 38 3603 

40 

8870108 

30 

8856639 

20 

8843095 

62° 

10' 

0.8829476 

62° 

0' 

8815782 

60 

8802014 

40 

8788171 

30 

8774254 

20 

8760263 

61° 

10' 

0.8746197 

6I0 

0' 

8732068 

60 

8717844 

40 

8703657 

30 

8689196 

20 

8674762 

600 

10' 

Sine 

D. 

M. 

NATURAL  SINES  AND  TANGENTS 


153 


^dTIl 

Sine 

Tangent 

CotHiii^ent 

Cosine 

/  D.  M. 

30°  0 

0.500000n 

0.5773503 

1.7o2U508 

0.8660254 

60^  0' 

10 

5025170 

5812353 

1.7204736 

8645673 

50; 

20 

5050298 

5851335 

1.70.90116 

8631019 

40 

30 

50763S4 

6890450 

1.6976631 

8616292 

30 

40 

5100426 

5929699 

1.6864261 

8601491 

20 

l30°  60' 

i 

6125425 

5969084 

1.6752988 

8586619 

59°  10' 

1 

131^  0' 

0.5150381 

0.6008606 

1.6642795 

0.8571673 

59°  0' 

10 

5175293 

604826C 

1.6533663 

8556655 

50 

20 

6200161 

6088067 

1.6425576 

8541564 

40 

30 

6224986 

6128008 

1.6318517 

8526402 

30 

40 

5249766 

6168092 

1.6212469 

8511167 

20 

31°  50' 

5274602 

6208320 

1.6107417 

8495860 

58°  10' 

32'>  0' 

0.5299193 

0.6248694 

1.6003345 

0.8480481 

58°  C 

10 

5323839 

62892 '4 

1.5900238 

84650.30 

50 

20 

5348440 

6329883 

1.5798079 

8449508 

40 

30 

6372996 

6370703 

1.5696856 

843  3914 

30 

40 

6397507 

6411673 

1.5596552 

8418249 

20 

3^°  50' 

5421971 

6452797 

1.5197155 

8402513 

67°  10' 

33^^  0' 

0.5446390 

0.6494076 

1.5398650 

0.8386706 

57°  0' 

10 

6470763 

633651 1 

1.5301023 

8>.70827 

50 

20 

6495090 

6577103 

1.5204261 

8354878 

40 

30 

5519370 

6618S56 

1.5108352 

8338858 

30 

40 

5543603 

6660169 

1.501.3282 

8322768 

20 

33°  50' 

5567790 

6702843 

1.4919039 

8306607 

56°  10' 

34^  0' 

0.5591929 

0.6745035 

1.4825610 

0.329037G 

56°  0' 

10 

6616021 

6737492 

1.4732933 

8274074 

50 

20 

5640066 

6830066 

1.4641147 

8257703 

40 

30 

5664062 

6872310 

1.4550090 

8241262 

30 

40 

668801 1 

6915725 

1.4459801 

8224751 

20 

34°  50' 

6711912 

G958813 

1.4370268 

8208170 

55''   10' 

1 

35°  0' 

0.5735764 

0. '700207  5 

1.4281480 

0.8191520 

55°  0' 

10 

6759568 

7045515 

1.4193427 

8171801 

50 

20 

6783323  | 

7089133 

1.4106098 

8158013 

40 

30 

5807030 

7132931 

1.4019483 

8141155 

30 

40 

5830687 

7176911 

1.3933571 

8124220 

20 

35°  50' 

5854294 

7221075 
Cotangent 

1.3848353 

8107234 

54°  10 
D.  -VI  ^ 

D.  M. 

Cosine 

Tangent  . 

Sine 

^1 


I  FA 


NATURAL  SINES  AND  TANGENTS. 


D.  m: 

36°  0' 

Sine 

Tnngent. 

0.7265425 

.Cotanger^t 
1.3763819 

Cosine 

U.M. 

0.5877853 

0.8090170 

54°  0' 

1    ]0 

5901361 

730f>963 

1.3679959 

8073038 

50 

t    20 

5924819 

7354691 

1.3596764 

8055837 

40 

30 

5948228 

7399611 

1.3514224 

8038569 

30 

40' 

5971586 

7444724 

1.3432331 

8021232 

20 

360  50 

5094893 

7490033 

1.3351075 

8003827 

630  10' 

=37°  0' 

0.6U18150 

0.7535541 

1.3270448 

0.7986355 

53°  C 

1    10 

6041356 

7581248 

1.3190441 

7968815 

50 

;   20 

6064511 

7627157 

1.3111046 

7951208 

40 

1    30 

6087614 

7673270 

1.3032254 

7933533 

30' 

40 

6110666 

7719589 

1.2954057 

7915792 

20 

37°  50' 

6133666 

7766118 

1.2876447 

78979&3 

52°  10' 

38°  0' 

0.6156615 

0.7812856 

1.2799416 

.0.7880108 

52°  0' 

10 

6179511 

7859808 

1.2722957 

7862165 

50 

20 

6202355 

7906975 

1.2647062 

7844157 

40 

30 

6225146 

7954359 

1.2571723 

7826082 

30 

40 

6247885 

8001963 

1.2496933 

7807940 

20 

38°  50' 

6270571 

8049790 

1.2422685 

7789733 

61°  10' 

i39°  0' 

06293204 

0.8097840 

1.2348972 

0.7771460 

51°  0' 

1    10 

6315784 

8146118 

1.2275786 

7753121 

50 

20 

6338310 

8194625 

1.2203121 

7734716 

40 

30 

6360782 

8243364 

1.2130970 

7716246 

30 

40 

6383201 

8292337 

1.2059327 

7697710 

20 

39°  50 

6405566 

8341547 

1.1.988184 

7679110 

50°  10' 

40°  0' 

0.6427876 

0.8390996 

1.1917536 

0.7660444 

50°  0' 

10 

6450132 

8440688 

1.1847370 

7641714 

50 

20 

6472334 

8490624 

1.1777698 

7622919 

40 

30 

6494480 

8540807 

1.170S496 

7604060 

30 

40 

6516572 

8591240 

1.1639763 

7585136 

20 

40°  50' 

6538609 

8641926 

1.1571495 

7566148 

49°  10' 

41°  0 

0  6560590 

0.8692867 

1.1503684 

0.7547096 

49°  0' 

10 

6582516 

8744067 

1.1436326 

7527980 

50 

20 

66043'66 

8795528 

1.1369414 

750f.800 

40 

30 

6626200 

8847253 

1.1302944 

7489557 

30 

40 

6647959 

8899244 

1.1236909 

7470251 

20  1 

41°  50' 

666~J66\ 

8951506 
Cotangent 

1.1171305 

7450881 

48°  lO'j 

D.  M.  ' 

D.  M. 

Cosine 

Tangent 

Sine 

NATURAL  SINES  AND  TANGENTS. 


155 


D.  M. 

Sine 

Tanjjent 

Cotangent 

Cosine 

D.  M. 

42°  0' 

0.6691306 

0.9004040 

1.1106125 

0.7431448 

48°  0' 

10 

6712895 

9056851 

1.1041365 

7411953 

60 

20 

6734427 

9109940 

1.0977020 

7392394 

40 

30 

6755902 

91b3312 

1.0913085 

7372773 

30 

^    40 

6777320 

9216969 

1.0849554 

7353090 

20 

42°  50' 

6798681 

9270914 

1.0786423 

7333345 

47°  10 

43°  0' 

0.6819984 

0.932-5151 

1.0723687 

0.7313537 

47°  0' 

10 

6841229 

9379683 

1.0661341 

7293668 

50 

20 

6862416 

9434513 

1.0599381 

7273736 

,  40 

30 

6883546 

9489646 

1.0537801 

7253744 

30 

40 

6904617 

9545083 

1.0476598 

7233690 

20 

43°  50' 

6925630 

9600829 

1.0415767 

7213574 

46°  10' 

44°  0' 

0.6946584 

0.9656888 

1.0355303 

0.7193398 

46°  0' 

10 

6967479 

9713262 

1.0295203 

7173161 

50 

20 

6988315 

9769956 

1.0235461 

7152863 

40 

30 

7009093 

9826973 

1.0176074 

7132504 

30 

40 

7029811 

9884316 

1.0117038 

7112086 

20 

44°  50' 

7050469 

9941991 

1.0058348 

7091607 

45°  10' 

45°  0'  0.7071068 

1.0000000 
Cotangent 

1.0000000 

0.70710G8 

45®  0' 

a  M. 

1 

D.  M. ' 

Cosine 

Tangent  ' 

Sine 

The  Secants  and  Cosecants^  which  are  not  inserted  in  this 
table,  may  be  easily  supplied.  If  1  be  divided  by  the  cosine 
of  an  arc,  the  quotient  will  be  the  secant  of  that  arc.  (Art. 
228.)  And  if  1  be  divided  by  the  «ine.  the  quotient  will  be 
fhe  cosecant. 


PI.  I. 


y.^cflin  ScM.ff. 


THICOWOMETHY. 


TRIGOXOMETRir, 


ri.ii. 


,/.»  Si.J^H. 


TRIOOITOSI GTR  V. 


^JuA    I „->^ 


^\ 


T«IGO:^©>tBTltY, 


i^m. 


TBICOyO>tKT»Y. 


r^ 


A  PRACTICAL  APPUCATION 

OP 

THE  PRINCIPLES  OF  GEOMETRY 

TO  THE 


OF 


SUPERFICIES  AND   SOLIDS 


Buiva 
THE  THIRD  PART 

OF 


A  COURSE  OF  MATHEMATICS, 

ADAPTED  TO  THE  METHOD  OF  INSTRUCTION  IN 
THS  AMERICAN  COLLEGES. 


By  JEREMIAH  DAY,  D.  D.  LL.  D. 

president  of  Yale  College, 
THE  SECONB  EDITION. 


NEW-HAVEN  : 

PRINTED  AND  PUBLISHED  BT  S.  CONTERSE' 

1B25. 


District  of  Connecticuti  ss. 

BE  IT  REMEMBERED,  That  on  the  second  day  of  March,  in  tht 
fortieth  year  of  the  Independence  of  the  United  States  of  America,  JER- 
EMIAH DAY,  of  the  said  District,  hath  tleposited  in  this  Office  the 
title  of  a  look,  the  right  whereof  he  claims  as  Author,  in  the  words 
following,  to  wit  :— 

..       T'Z^  "  A  practical  Application  of  the  Principles  of  Geometry  to  the  Mensu* 

ration  of  Superficies  and  Solids  :  Being  the  Third  Part  of  a  Course  of  Mathematics,  adapted 
to  the  method  of  instruction  in  the  American  Colleges.  By  Jeremiah  Day,  Professor  of 
Mathematics  and  ITatunl  Philosophy  in  Yale  College." 

In  conformity  to  the  act  of  the  Congress  of  the  United  States,  enUtled  "  An  act  for  the 
encouragement  of  learning,  by  securing  the  copi«R  of  Maps,  Charts,  and  Books,  to  the  authers 
»nd  proprietors  of  such  copies,  during  the  times  therein  mentioned  " 

HENRY  W    EDWARDS,  Clerk  of  tht  IHHrict  of  Conneetieut 

A  true  copy  of  Record,  examined  by  ifte, 

HENRY  W.  EDWARDS,  Cltrk  of  th$  Dittrict  o/  Connecticut 


Thk  following  short  Treatise  contains  little  more  than  an 
application  of  the  principles  of  Geometry,  to  the  numerical 
calculation  of  the  superficial  and  solid  contents  of  such 
figures  as  are  treated  of  in  the  Elements  of  Euclid.  As  the  plan 
proposed  for  the.work  of  which  this  number  is  a  part,  does  not 
admit  of  introducing  rules  and  propositions  which  are  not 
demonstrated ;  tlie  particular  consideration  of  the  areas  of 
the  Conic  Sections  and  other  curves,  with  the  contents  of 
solids  produced  by  their  revolution,  is  reserved  for  succeed- 
ing parts  of  the  course.  The  student  would  be  Httle  profited 
by  applying  arithmetical  calculation,  in  a  mechanical  way,  to 
figures  of  which  he  has  not  yet  learned  even  the  definitions. 
But  as  this  number  may  fall  into  the  hands  of  some  who  will 
not  read  those  which  are  to  follow,  the  principal  rules  for 
conic  areas  and  solids,  and  for  the  gauging  of  casks,  are  given, 
without  demonstrations,  in  the  appendix.  Those  who  wish  to 
take  a  complete  view  of  Mensuration,  in  all  its  parts,  are  re- 
ferred to  the  valuable  treatise  of  Dr.  Hutton  on  the  subject. 


CONTENTS, 

P»ge. 

Section  I.     Areas  of  figures  bounded  by  right  lines,  1 

II.     The  Quadrature  of  the  circle  and  its  parts,  14 

Promiscuous  examples  of  Areas,             -  25 

HI.     Solids  bounded  by  plane  surfaces,           -  28 

IV.     The  Cylinder,  Cone,  and  Sphere,            -  43 

Promiscuous  examples  of  Solids,             -  59 

V.     Isoperimetry,     -             -            -             -  61 

APPENDIX.— Part   I. 

Mensuration  of  the  Conic  Sections,  and  other  figures,  72 

Part  II. 

Gauging  of  Casks,            ...            -  80 

Notes,     -..---  86 

Table  of  Circular  Segments,         -            -             -  93 


SECTION  I. 


AREAS  OF  FIGURES  BOUNDED  BY  RIGHT  LINES. 


-        J     T^HE  following  definitions,  which  are  nearly  the 
^  '    '    -^    same  as  in  Euclid,  are  inserted  here  for  the  con- 
venience of  reference. 

1.  Four-sided  figures  have  different  names,  according  to  the 
relative  position  and  length  of  the  sides.  A  parallelogram  has 
its  opposite  sides  equal  and  parallel ;  as  ABCD.  (Fig.  2.)  A 
rectangle,  or  right  parallelogram,  has  its  opposite  sides  equal, 
and  all  its  angles  right  angles;  as  AC.  (Fig  1.)  A  square 
has  all  its  sides  equal,  and  all  its  angles  right  angles ;  as  AB 
GH.  (Fig  3  )  A  rhombus  has  all  its  sides  equal,  and  its  an- 
gles oblique:  as  ABCU.  (Fig.  3.)  A  rhomboid  has  its  op- 
posite sides  equal,  and  its  angles  oblique  ;  as  ABCD.  (Fig. 
2.)  A  trapezoid  has  only  two  of  its  sides  parallel ;  as  ABLD. 
(Fig.  4  )     Any  other  four  sided  figure  is  called  a  trapezium. 

II.  A  figure  which  has  more  than  four  sides  is  called  a 
polygon.  A  regular  polygon  has  all  its  sides  equal,  and  all 
its  angles  equal. 

III.  The  height  of  a  triangle  is  the  length  of  a  perpendicu- 
lar, drawn  from  one  of  the  angles  to  the  opposite  side;  as 
CP.  (Fig.  5.)  The  height  oi  a  four-sided  figure  is  the  per- 
pendicular distance  between  two  of  its  parallel  sides;  as  CP. 
(Fig.  4.) 

IV.  The  area  or  superficial  contents  of  a  figure  is  the  space 
contained  within  the  line  or  lines  by  which  the  figure  is 
bounded. 

2.  In  calculating  areas,  some  particular  portion  of  surface 
is  fixed  upon,  as  the  measuring  unit,  with  which  the  given 
figure  is  to  be  compared.  This  is  commonly  a  square;  as  a 
square  inch,  a  square  foot,  a  square  rod,  he.  For  this  rea- 
son, determining  the  quantity  of  surface  in  a  iigure  is  called 
squaring  it,  or  finding  its  quadrature  ;  that  is,  finding  a  square 
or  number  of  squares  to  which  it  is  equal. 

2 


^  MENSURATION  OF 

3.  The  superficial  unit  has  generally  the  same  name,  as 
the  linear  un'\t  which  forms  the  side  of  the  square. 

The  side  of  a  square  inch  is  a  linear  inch  ; 
of  a  square  foot,  a  linear  foot ; 
of  a  square  rod,  a  linear  rod,  &;c. 

There  are  some  superficial  measures,  however,  which  have 
no  corresponding  denominations  of  length.  The  acre,  for 
instance,  is  not  a  square  which  has  a  line  of  the  same  name 
for  its  side. 

The  following  tables  contain  the  linear  measures  in  com- 
mon use,  with  their  corresponding  square  measures. 

Linear  Measures,  Square  Measures. 


12 

Inches 

=  1  Foot.            144 

Inches 

=  1  Foot. 

3 

Feet 

=  1  Yard.                9 

Feet 

=  1  Yard- 

6 

Feet 

=  1  Fathom.         36 

Feet 

=  1  Fathom. 

161 

Feet 

=  1   Rod.             2721 

Feet 

=  1  Rod. 

51 

Yards 

=  1  Rod.               301 

Yards 

=  1  Rod. 

4 

Rods 

=  1   Chain.             16 

Rods 

=  1   Chain. 

40 

Rods 

=  1   Furlong.     1600 

Rods 

=  1   Furlong. 

320 

Rods 

=  1   Mile.       102400 

Rods 

=  1   Mile. 

An  acre  contains  160  square  rods,  or  10  square  chains. 
By  reducing  the  denominations  of  square  measure,  it  will 
be  seen  that 

1  sq.  mile=640  ;icre?=102400  rods=27a7n-100  ft.=401 4409600  inche?. 
1  acre=10  chains=l60  rods=43560  feet^627^2640  inches. 

The  fundamental  problem  in  the  mensuration  of  superfi- 
cies is  the  very  simple  one  of  determining  the  area  of  a  right 
parallelogram.  The  contents  of  other  figures,  particularly 
those  which  are  rectilinear,  may  be  obtained  by  finding  par- 
allelograms which  are  equal  to  them,  according  to  the  princi- 
ples laid  down  in  Euclid. 

PROBLEM     I. 

To  find  the  area  of  a  parali.elogram,  square,  rhombus,  or 
rhomboid, 

4.  MULTIPLY  THE  LENGTH  BY  THE  PERPENDICULAR  HEIGHT  OR 
BREADTH. 

It  is  evident  that  the  number  of  square  inches  in  the  par- 
allelogram AC  (Fig.  1 .)  is  equal  to  the  number  of  linear  inches 
in  the    length    AB,  repealed   as  many  times  as   there  are 


PLANE  SURFACES.  3 

inches  in  the  breadth  BC.    For  a  more  particular  illustration 
of  this,  see  Alg.  511 — 514. 

The  oblique  parallelogram  or  rhomboid  ABCD  (Fig.  2.) 
is  equal  to  the  right  parallelogram  GHCD.  (Euc.  36.  1.) 
The  area,  therefore,  is  equal  to  the  length  AB  multiplied 
into  the  perpendicular  height  HC.  And  the  rhombus  ABCD 
{Fig.  3.)  is  equal  to  the  parallelogram  ABGH.  As  the  sides 
of  a  square  are  all  equal,  its  area  is  found,  by  mulliplying  out 
of  the  sides  into  itself, 

Ex.  1 .  How  many  square  feet  are  there  in  a  floor  23^  feet 
long,  and  18  feet  broad  ?  Ans.  23^X18=423. 

2.  What  are  the  contents  of  a  piece  of  ground  which  is  66 
feet  square  ?  Ans.  4356  sq.  feet=16  sq.  rods. 

3.  How  many  square  feet  are  there  in  the  four  sides  of  a 
room  which  is  22  feet  long,  17  feet  broad,  and  11  feet  high  ? 

Ans.  858. 

Art.  5.  If  the  sides  and  angles  of  a  parallelogram  are  given, 
the  perpendicular  height  may  be  easily  found  by  trigonome- 
try. Thus  CH  (Fig.  2.)  is  the  perpendicular  of  a  right 
angled  triangle,  of  which  BC  is  the  hypothenuse.  Then 
(Trig.  134.) 

R  :  BC::SinB  :CH. 

The  area  is  obtained  by  multiplying  CH  thus  found,  into 
the  length  AB. 

Or,  to  reduce  the  two  operations  to  one, 

As  radius, 

To  the  sine  of  any  angle  of  a  parallelogram  ; 

So  is  the  product  of  the  sides  including  that  angle, 

To  the  area  of  the  parallelogram. 

For  the  arca=AB  XCH  (Fig.  2.)  But  CH=JP^^'"-g 

R 

Therefore, 
The  area  =  ^^^^^"^^'"^  OrR  :  SinB::AB  X  BC  :  the  area. 

Ex.  If  the  side  AB  be  58  rods,  BC  42  rods,  and  the  angle 
B  63*,  what  is  the  area  of  the  parallelogram  ? 


MENSURATION  OF 


As  radius 

10.00000 

To  the  sine  of  B 

63» 

9.94988 

So  is  the  product  of  AB 

58 

1.76343 

Into  BC  (Trig.  39,) 

42 

1.62325 

To  the  area  2170.5  sq.  rods  3.3365G 

2.  If  the  side  of  a  rhombus  is  67  feet,  and  one  of  the 
angles  73^  what  is  the  area  ?  Ans.  4292.7  feet. 

6.  When  the  dimensions  are  given  in  feet  and  inches,  the 
multiplication  may  be  conveniently  performed  by  the  arith- 
metical T[i\e  o(  Duodecimals;  in  which  each  inferior  denom- 
ination is  one  twelfth  of  the  next  higher.  Considering  a  foot 
as  the  measuring  unit^  a  prime  is  the  twelfth  part  of  a  foot ; 
a  second,  the  twelfth  part  of  a  prime,  &c.  It  is  to  be  ob- 
served, that,  in  measures  of /en^//i,  inches  ^re primes;  but  in 
superficial  measure  they  are  seconds*  In  both,  a  prime  is  jj 
of  a  foot.  But  j\  of  a  square  foot  is  a  parallelogram  a  foot 
long  and  an  inch  broad.  The  twelfth  part  of  this  is  a  square 
inch,  which  is  j^j  of  a  square  foot. 

Ex.  1.  What  is  the  surface  of  a  board  9  feet  5  inches,  by 
2  feet  7  inches. 


F 

9 

5' 

2 

7 

18 

10 

5 

5 

11 

24 

3 

ir 

11",  or  24  feet  47  inches. 

2.  How  many  feet  of  glass  are  there  in  a  window  4  feet  11 
inches  high,  and  3  feet  5  inches  broad  ? 

Ans.  16F.  9'  7",  or  16  feet  115  inches. 

7.  If  the  area  and  one  side  of  a  parallelogram  be  given,  the 
other  side  may  be  found  by  dividing  the  area  by  the  given  side. 
And  if  the  area  of  a  square  be  given,  the  side  may  be  found 
by  extracting  t'e  square  root  of  the  area.  This  is  merely  re- 
versing the  rule  in  art.  4.     See  Alg.  520,  521. 

Ex.  1.  What  is  the  breadth  of  a  piece  of  cloth  which  is  36 
yds.  long,  and  which  contains  63  square  yds.    Ans.  If  yds. 


PLANE  SURFACES.  5 

2.  What  is  the  side  of  a  square  piece  of  land  containing 
289  square  rods  ? 

3.  How  many  yards  of  carpeting  \\  yard  wide,  will  cover 
a  floor  30  feet  long  and  221  broad  ? 

Ans.  30X22|feet  =  10x7i=75yds.  And  75-f.li=60. 

4.  What  is  the  side  of  a  square  which  is  equal  to  a  paral- 
lelogram 936  feet  long  and  104  broad  ? 

5.  How  many  panes  ®f  8  by  10  glass  are  there,  in  a  win- 
dow 5  feet  high,  and  2  feet  8  inches  broad  ? 

PROBLEM    11. 

To  find  the  area  of  a  t&ianole. 

8.  Rule  I.  Multiply  one  side  nv  half  the  perpen- 
dicular FROM  THE  OPPOSITE  ANGLE.  Op,  multiply  half  the 
side  by  the  perpendicular.  Or,  multiply  the  whole  side  by 
the  perpendicular,  and  lake  half  the  product. 

The  area  of  the  triangle  ABC  (Fig.  5.)  is  equal  to  |  PCx 
AB,  because  a  parallelogram  of  the  same  base  and  height  is 
equal  to  PCxAB,  (Art.  4.)  and  by  Euc.  41.  1,  the  triangle 
is  half  the  parallelogram. 

Ex.  1.  If  AB  (Fig.  5.)  be  65  ie^t,  and  PC  31.2,  what  is 
the  area  of  the  triangle  ?  Ans.  1014  square  ieeU 

2.  What  is  the  surface  of  a  triangular  board,  whose  base 
is  3  (eti  2  inches,  and  perpendicular  height  2  feet  9  inches? 
Ans.  4F.  4'  3'',  or  4  feet  51  inches. 

9.  If  two  sides  of  a  triangle  and  the  included  angle,  are 
given,  the  perpendicular  on  one  of  these  sides  may  be  easily 
found  by  rectangular  trigonometry.  And  the  area  may  be 
calculated  in  the  same  manner  as  the  area  of  a  parallelogram 
in  art.  5.     In  the  triangle  ABC  (Fig.  2  ) 

R:BC::SinB  :CH 
And  because  the  triangle  is  half  the  parallelogram  of  the 
same  base  and  height, 
As  radius, 

To  the  sine  of  any  angle  of  a  triangle  ; 
So  is  the  product  of  the  sides  including  that  angle, 
To  twice  the  area  of  the  triangle.     ( Art.  5.) 
Ex.   If  AC  (Fig.  5.)  be  39  feet,  AB  G5  feet,  and  the  angle 
at  A  53°  7'  48",  what  is  the  area  of  the  triangle  ? 

Ans.  1014  square  feet. 


6  MENSURATION  OF 

9.  b.  \ione  side  and  the  angles  are  given  ;  then 

As  the  product  of  radius  and  the  sine  of  the  angle  opposite 
the  given  side, 

To  the  product  of  the  sines  of  the  two  other  angles ; 
So  is  the  square  of  the  given  side, 
To  twice  the  area  of  the  triangle. 

If  PC  (Fig.  5.)  be  perpendicular  to  AB. 

R  :  Sin  B::BC  :CP 

Sin  ACB  :  Sin  A::AB  :  BC 
Therefore  (Alg.  390,  382.) 
R  X  Sin  ACB  :  Sin  A  X  Sin  B  : :  AB  X  BC  :  CPxBC:: 
AB^  :  ABxCP= twice  the  area  of  the  triangle. 

Ex  If  one  side  of  a  triangle  be  57  feet,  and  the  angles  at 
the  endsof  this  side  50°  and  60°,  what  is  the  area  ? 

Ans.  1 147  sq.  feet. 

10.  If  the  sides  only  of  a  triangle  are  given,  an  angle  may 
be  found,  by  oblique  trigonometry,  Case  IV,  and  then  the 
perpendicular  and  the  area  may  be  calculated.  But  the  area 
may  be  more  directly  obtained,  by  the  following  method. 

Rule  II.  When  the  three  sides  are  given,  from  half  their 
sum  subtract  each  side  severally  ^multiply  together  the  half  sum 
and  the  three  remainders,  and  extract  the  square  root  of  the 
product. 

If  the  sides  of  the  triangle  are  a,  h,  and  c,  and  if  A=half 
their  sum,  then 

The  area=\/hX{h'-a)X{h  —  b)X{h-c) 

For  the  demonstration  of  this  rule,  see  Trigonometry, 
Art.  221. 

If  the  calculation  be  made  by  logarithms,  add  the  loga- 
rithms of  the  several  factors,  and  half  their  sum  will  be  the 
logarithm  of  the  area.     (Trig.  39,  47.) 

Ex.  I.  In  the  triangle  ABC  (Fig.  5.)  given  the  sides  a  52 
feet,  b  39,  and  c  65 ;  to  find  the  side  of  a  square  which  has 
the  same  area  as  the  triangle. 

^(a+b-\'c)^h=7Q  A- 6  =  39 

h'-a=2G  A  — c=13 

Then  the  area=v'78  X26  X39  X  13  =  1014  square  feet. 


PLANE  SURFACES. 

By  logarithms. 

The  half  sum  =78  1.89209 

First  remainder  =26  1.41497 

Second  do.  =39  1.69106 

Third    do.  =13  1.11394 


2)6.01206 

The  area  required        =1014  2)3.00603 

Side  of  the  square    =31.843  (Trig.  47)       1.60301 

2.  If  the  sides  of  a  triangle  are  134,  108,  and  80  rods,  what 
is  the  area  ?  Ans.  4319. 

3.  What  is  the  area  of  a  triangle  whose  sides  are  37  J ,  264, 
and  225  feet  ? 

11.  In  an  equilateral  triangle,  one  of  whose  sides  is  a,  the 
expression  for  the  area  becomes 

^hx{h-a)x{h—a)X(h-a) 

But  as  h  =  ^a,  and  h  —  a  =  ^a  —  a  =  laf  the  area  is 

y/laX^aX^aXia  =  y/j\a'=laW3  (Alg.  271.) 

That  is,  the  area  of  an  equilateral  triangle  is  equal  to  i  the 

square  of  one  of  its  sides,  multiplied  into  the  square  root  of  3, 

which  is  1.732. 

Ex.  1.  What  is  the  area  of  a  triangle  whose  sides  are 
each  34  feet.'*  Ans.  500^  feet. 

2.  If  the  sidesof  a  triangular  field  are  each  100  rods,  how 
manf  acres  does  it  contain  .'' 

PROBLEM    III. 

To  find  the  area  of  a  trapezoid. 

21.  Multiply  half  the  sum  of  the  parallel  sides  into  their 
perpendicular  distance. 

The  area  of  the  trapezoid  ABCD  (Fig.  4.)  is  equal  to  half 
the  sum  of  the  sides  AB  and  CD,  multiplied  into  the  perpen- 
dicular distance  PC  or  AH.  For  the  whole  figure  is  made 
up  of  the  two  triangles  ABC  and  ADC  ;  the  area  of  the  first 
of  which  is  equal  to  the  product  of  half  the  base  AB  into 
the  perpendicular  PC,  (Art,  8.)  and  the  area  of  the  other  is 
equal  to  the  product  of  half  the  base  DC  into  the  perpendic- 
ular AH  or  PC. 


B  MENSURATION  OF 

Ex.  If  AB  (Fig.  4.)  be  46  feet,  BC  31,  DC  38,  and  the 
angle  B  70®,  what  is  the  area  of  the  trapezoid  ? 
R  :  BC::SinB  :  PC=29.13.     And  42x29.13=1223^. 

2.  What  are  the  contents  of  a  field  which  has  two  parallel 
sides  65  and  38  rods,  distant  from  each  other  27  rods  P 


PROBLEM    IT. 

To  find  the  area  of  a  trapezium,  or  of  an  irregular  polygon. 

13.  Divide  the  whole  figure  into  triangles,  by  drawing 
diagonals,  and  find  the  sum  of  the  areas  of  these  triangles. 
(Alg.519.) 

If  the  perpendiculars  in  two  triangles  fall  upon  the  same  di- 
agonal, the  area  of  the  trapezium  formed  of  the  two  trian- 
gles, is  equal  to  half  the  product  of  the  diagonal  into  the  sum 
of  the  perpendiculars. 

Thus  the  area  of  the  trapezium  ABCH  (Fig.  6.)  is 
iBHxAL-f-iBHxCM=iBHx(AL4-CM) 
Ex.     In  the  irregular  poloygon  ABCDH  (Fig.  6.) 

CBH  =  S6  I  ALi=5.3 

if  the  diagonals  <  r^rj     oc>'  and  the  perpendiculars  <  CM=9.4 

^CH  =  3J,  r  (DN=7.3 

The  area  =  18x  14.6  +  16x7.3=379.6 

14.  If  the  diagonals  of  a  trapezium^re  given,  the  area  may 
be  found,  nearly  in  the  same  manner  as  the  area  of  a  paral- 
lelogram in  Art.  5,  and  the  area  of  a  triangle  in  Art.  9. 

In  the  trapezium  ABCD  (Fig.  8.)  the  sines  of  the  four  an- 
gles at  N,  the  point  of  intersection  of  the  diagonals,  are  all 
equal.  For  the  two  acute  angles  are  supplements  of  the  other 
two,  and  therefore  have  the  same  sme.  (Trig.  90.)  Put- 
ting, then,  Sin  N  for  the  sine  of  each  of  these  anglps,  the 
areas  of  the  four  triangles  ol  which  the  trapezium  is  com- 
posed, are  given  by  the  following  proportions;  (Art.  9.) 

fBNxAN  :  2  area  ABN 
n.Q-    r<r.JBNxCN  t2arenBCN 

n.^inl\..<^^  XCN  :  2  area  CDN 

I  DNxAN:  2  area  ADN 


PLANE  SURFACES.  9 

And  by  addition,  (Alg.  388,  Cor.  1.*) 

R:SinN::BNxAN-fBNxCN-|-DNxCN+DNxAN:2 
area  ABCD. 

The  3d  term=(AN+CN)x(BN-f-DN)=ACxBD,  by 
the  figure. 

Therefore,  R  :  Sin  N : :  AC  X  BD  :  2  area  ABCD.  That  is, 

As  Radius, 

To  the  sine  of  the  angle,  at  the  intersection  of  the  di- 
agonals of  a  trapezium  ; 
So  is  the  product  of  the  diagonals, 
To  twice  the  area  of  the  trapezium. 

It  is  evident  that  this  rule  is  applicable  to  a  parallelogram, 
as  well  as  to  a  trapezium. 

If  the  diagonals  intersect  at  right  angles^  the  sine  of  N  is 
equal  to  radius;  (Trig.  95.)  and  therefore  the  product  of  the 
diagonals  is  equal  to  twice  the  area.     (Alg.  395.  t) 

Ex.  1.  If  the  two  diagonals  of  a  trapezium  are  37  and  62, 
and  if  they  intersect  at  an  angle  of  54°,  what  is  the  area  of 
the  trapezium  ?  Ans.  928. 

2.  If  the  diagonals  are  85  and  93,  and  the  angle  of  inter- 
section, 74°,  what  is  the  area  of  the  trapezium  ? 

14.  b.  When  a  trapezium  can  be  inscribed  in  a  circle^  the 
area  may  be  found  by  either  of  the  following  rules. 

I.  Multiply  together  any  two  adjacent  sides^  and  also  the 
two  other  sides ;  then  multiply  half  the  sum  of  thesi  products 
by  the  sine  of  the  angle  included  by  either  of  the  pairs  of  sides 
multiplied  together. 

Or, 

II.  From  half  the  sum  of  all  the  sides,  subtract  each  side 
severally,  multiply  together  the  half  sum  and  the  four  remain- 
ders, and  extract  the  square  root  of  (he  product. 

If  the  sides  are  a,  b,  c,  and  d  ;  and  if  /i=half  their  sum  ; 


The  area  =  ^{h- a)x{h-b)X(h—c)X{h  —  d) 

*  |:uclid  2,  5.  Cor.  t  Euc.  14.  5. 

3 


10  MENSURATION  OF 

If  the  trapezium  ABCD  (Fig.  33  )  can  be  inscribed  in  a 
circle,  the  sum  of  the  opposite  angles  BAD  and  BCD  is  180° 
(Euc.  22.  3  )     Tlierefore  the  sine  of  BAD  is  equal  to  that  of 
BCD  or  PCD. 
If  5=the  sine  of  cither  of  these  angles,  radius  being  1 ,  and  if 

AB=a,     BC=b,     CD=c,     AD=f/; 

The  triangle  BAD=^adXs,     And  BCD  =  Uc%s;  (Art.  9.) 

Therefore, 

I.    Theareaof\BCD  =  ^{ad+bc)Xs. 

To  obtain  the  value  of  s,  in  terms  of  the  sides  of  the  tra- 
pezium, draw  DP  and  DP'  perpendicular  to  BA  and  BC. 
Then        Rad.  :5::AD:DP::CD:DP'. 
Also  AP==AD^  -DP-%and  CP'^_=CD^  -DP^'- 

So  that^j)p^QjQ^^.^^,Ana  ^(jp^^^._,.,3  =,^^1_,. 

C  BP  =AB-A?=a—dx/'  -s^ 
But  by  the  figure  [  b?'==BC  +  CP' =b+cx/r^^ 


And  BP'-fDP'=DB'=BP'+DP 


That  is  a^  -2adx/l  - .r-  +d^  =b''  '{-'2bc^/ \  -s-  +€'' 
Rcvlucing  the  equation,  we  have 

5«  =  1  —  ^1 ! 1_,  and 

(2ad-{'2bcy        ' 


V(2a(/4-26c)=^  -  (b-  -hc^-a''  -  d'-)'- 
'  '2ad-\-2bc 


Substituiing  for  s  in  the  first  rule,  the  value  here  found,  we 
have  the  area  of  the  trapezium,  equal  to 


\^{:^ad4-2bcy  -  (62  4-c2  -a^-d^y 

The  expression  under  the  radical  sign  is  the  difference  of 
two  squares,  and  may  be  resolved,  as  in  Trig.  221,  into  the 
factors  

(b^c'^a~d'  x(a-{-d'  -b-c') 
and  these  again  into 
{a'{-b+c-d){b  +  c+d-a){a'{-b-{-d-c){a-\-d-i'C-b) 

Jfthen  A=halflhe  sum  of  the  sides  of  the  trapezium, 


PLANE  SURFACES.  11 

11.   The  area  =  ^^{h-a)X(h-b)X(h-c)X(k-d) 
If  on»  .J    the  sides,  as  d,  is  supposed  to  be  diminished,  tiltHt 
is  ro-iuced  to  nothing  ;  the  figure  becomes  a  triangle^  and  the 
expression  tor  the  area  is  the  same  as  in  art.  10.     See  Hut- 
ton's  Mensuration. 

PROBLEM    V. 

To  find  the  area  of  a  regular  polygon. 

1 5.  Multiply  one  of  its  sides  into  half  its  perpendicular  dis- 
tance FROM  THE  CENTRE,  AND  THIS  PRODUCT  INTO  THE  NUMBER  OF  SIDES. 

A  regular  polygon  contains  as  many  equal  triangles  as  the 
figure  has  sides.  Thus  the  hexa^ion  ABDFGH  (Fig.  7.) 
contains  six  triangles,  each  equal  to  ABC.  The  area  of  one 
of  them  is  equ-^l  to  the  product  of  the  side  AB,  into  half  the 
perpendicular  CP  (Art.  8.)  The  area  of  the  whole,  there- 
fore, is  equal  to  this  product  multiplied  into  the  number  oi 
sides. 

Ex.1.  What  is  the  area  of  a  regular  octagon,  in  which 
the  lenu^th  of  a  side  is  60,  and  the  perpendicular  from  the 
centre  7^2. 426  ?  Ans.  17.582. 

2.  What  is  the  area  of  a  regular  decagon  whose  sides  are 
46  each,  and  the  perpen<i'cular  70.7867  ? 

16.  If  only  the  length  and  number  of  sides  of  a  regular 
polyiion  be  given,  ihe  perpendicular  from  the  centre  may  be 
easily  found  bj  trigonometry.  The  periphery  of  the  circle 
in  which  the  polygon  is  mscribed,  is  divided  into  as  many 
equal  pans  as  the  polygon  has  sides.  (Euc.  16.  4.  Schol.) 
The  arc,  of  which  one  of  the  sides  is  a  chord,  is  therefore 
knowt)  :  and  of  course,  the  angle  at  the  centre  subtended  by 
this  arc. 

Lei  AB  (Fig.  7.)  be  one  side  of  a  regular  polygon,  in- 
scribed in  the  circle  ABDG.  The  perpendicular  CP  bisects 
the  line  AB,  and  the  angle  ACB.  (Euc.  3.  3.)  Therefore 
BCP  is  the  same  part  of  360°,  which  BP  is  of  the  perimeter 
of  the  polygon.  Then,  in  the  right  angled  triangle  BCP,  if 
BP  be  radius,  (Trig.  i22.) 

R  :  BP:  :Cotan  BCP  :  CP.      That  is, 


IJ  MENSURATION  OF 

As  Radius, 

To  half  of  one  of  the  sides  of  the  polygon  ; 
So  is  the  cotangent  of  the  opposite  angle, 
To  the  perpendicular  from  the  centre. 

Ex.  1.  If  the  side  of  a  regular  hexagon  (Fig.  7.)  be  38 
inches,  what  is  the  area? 

The  angle  BCP=t'2  of  360°=30°.     Then, 

R  :  I9::Cot30*  :  32.909=CP,  the  perpendicular. 

And  the  area=l9X32.909X6=3751.6. 

2.  What  is  the  area  of  a  regular  decagon  whose  sides  are 
each  62  feet  ?     Ans.  29576. 

17.  From  the  proportion  in  the  preceding  article,  a  lahlt 
of  perpendiculars  aftd  areas  maybe  easily  formed,  for  a  series 
of  polygons,  <if  which  each  side  is  a  unit.  Putting  R  =  l, 
(Trig.  100.)  and  n=:the  number  of  sides,  the  proportion  be- 
comes 

1  :  I : :  ^ot-^  :    the  perpendicular, 

r.       ,  ,  360 

bo  that,  me  »cr».  =  ±Cot-;r- 

And  the  area  is  equal  to  half  the  product  of  the  perpen- 
dicular into  the  number  of  sides.  (Ait.  15.) 

Thus,  in  the  trigon,  or  equilateral  triangle,  the  perpendicu- 
^    360° 
lar=J  Cot-g-=J  Cot  60°=0.2886752. 

And  the  area =0.4330127. 

»      ,  .  ,.     .  ^    360^ 

In  the  tetragon  or  square,  the  perpendicular =|  Cot— ^ 

=i  Cot.  45°=0.5.     And  the  area=l. 

In  this  manner,  the  following  table  is  formed,  in  which  the 
side  of  each  polygon  is  supposed  to  be  a  unit. 


PLANE  SURFACES. 


13 


A  TABLE  OF  REGULAR  POLYGONS. 


JS^ames. 


Sides. \Angles.\Perpendiculars  \      Areas. 


Trigon 

Tetragon 

Pentagon 

Hexagon 

Heptagon 

Octagon 

Nonagon 

Decagon 

Undecagonj 

Dodecagon 


3 

60» 

4 

45° 

5 

36° 

6 

30° 

7 

254 

8 

221 

9 

20° 

10 

18° 

11 

16t\ 

12 

15° 

0.2886752 
0.5000000 
0.6881910 
0.8660254 
1.0382601 
1.2071069 
1.3737385 
1.5388418 
1.7028439 
1.8660252 


0.4330127 
1.0000000 
1.7204774 
2.5980762 
3.6339124 
4.8284271 
6.1818242 
7.6942088 
9.3656399 
11.1961524 


By  this  table  may  be  calculated  the  area  of  any  other 
regular  polygon,  of  the  same  number  of  sides  with  one  of 
these.  For  the  areas  of  similar  polygons  are  as  the  squares  of 
their  homologous  sides.     (Euc.  20,  6.) 

To  find,  then,  the  area  of  a  regular  polygon,  multiply  the 
square  of  one  of  its  sides  by  the  area  of  a  similar  polygon  of 
which  the  side  is  a  unit. 

Ex.  1.  What  is  the  area  of  a  regular  decagon  whose  sides 
are  each  102  rods.^  Ans.  80050.5  rods. 

2.  What  is  the  area  of  a  regular  dodecagon  whose  sides 
are  each  87  feet  ^ 


SECTION   II.* 


THE  QUADRATURE  OF  THE  CIRCLE  AND  ITS  PARTS. 


A  ,o  rk  £  ^'  T  A  CIRCLE  IS  a  plane  bounded 
Art.  18.     Definition  \.  jhl  i         ,-         i-  u  •  u    a- 

'^  by  a  line  which  is  equally  dis- 

tant in  all  its  parts  from  a  point  within  called  the  centre. 
The  bounding  line  is  called  the  circumference  or  periphery. 
An  arc  is  any  portion  of  the  circumference.  A  semi-circle 
is  half,  and  a  quadrant  one-fourth,  of  a  circle. 

II.  A  Diameter  of  a  circle  is  a  straight  line  drawn  through 
the  centre,  and  terminated  both  ways  by  the  circumference. 
A  Radius  is  a  straight  line  extending  from  the  centre  to  the 
circumference.  A  Chord  is  a  straight  line  which  joins  the 
two  extremities  of  an  arc. 

III.  A  Circular  Sector  is  a  space  contained  between  an  arc 
and  the  two  radii  drawn  from  the  extremities  of  the  arc.  It 
may  be  less  than  a  semi-circle,  as  ACBO,  (Fig.  9.)  or  greater, 
as  ACBD. 

IV.  A  circular  Segment  is  the  space  contained  between  an 
arc  and  its  chord,  as  ABO  or  ABD.  (Fig.  9  )  The  chord 
is  sometimes  called  the  base  of  the  segment.  The  height  of 
a  segment  is  the  perpendicular  from  the  middle  of  the  base  to 
the  arc,  as  PO.  (Fiji.  9.) 

V.  A  Circular  Zone  is  the  space  between  two  parallel 
chords,  as  AGHB.  (Fig.  15.)  It  is  called  the  middle  zone, 
when  the  two  chord?  are  equal. 

VI.  A  Circular  Ring  is  the  space  between  the  peripheries 
of  two  concentric  circles,  as  AA'  BB'.  (Fig.  13.) 

VII.  A  Lune  or  Crescent  is  the  space  between  two  circu- 
lar arcs  which  intersect  each  other,  as  ACBD.  (Fig.  14.) 

19.  The  Squaring  of  fhe  Circ/<' is  a  problem  which  has  ex- 
ercised the   ingenuity  of  distinguished    mathematicians    for 

*  WalHs's  Algebra,  Le  Gendre's  Geometry,  Book  iv.  and  Note  iv.  Hut- 
ton's  Mensuration,  Horseley's  Tri.'onoraetry,  Book  i,  Sec.  3  ;  Introduction 
toEuler's  Analyaisof  Infinites,  London  Phil.  IVans.  Vol  vi  No.  75,  Lxvi.  p. 
476,  X.XXXIV.  p.  217,  and  Button's  abridgmeal  of  do.  Vol.  ii.  p.  S47. 


MENSURATION  OF  THE  CIRCLE.  15 

many  centuries.  The  result  of  their  efforts  has  been  onlj 
an  approximation  to  the  value  of  the  area.  This  can  be 
carrir-d  to  a  degree  of  exactness  far  beyond  what  is  necessary 
for  practical  purposes. 

20.  If  the  circumference  of  a  circle  of  given  diameter  were 
known,  its  area  could  be  easily  found.  For  the  area  is  equal 
to  the  product  of  hai!  the  circumference  into  half  the  diam- 
eter (Sup.  Euc  5,1.*)  But  the  circumference  of  a  circle 
has  never  heen  exactly  determined.  The  method  of  approx- 
imating to  it  is  by  inscribing  and  circumscribing  |?o/y^072s,  or 
by  some  process  of  calculation  which  is,  in  principle,  the 
same.  The  perimeters  of  the  polygons  can  be  easily  and 
exactly  determined.  That  which  is  circumscribed  isgreater, 
and  that  which  is  inscribed  is  less^  than  the  periphery  of  the 
circle;  and  by  increasing  the  number  of  sides,  the  difference 
of  the  two  polygons  may  be  made  less  than  any  given  quan- 
tity. (Sup.  Euc.  4,  1.) 

21  The  side  of  a  hexagon  inscribed  in  a  circle,  as  AB, 
(Fig.  7.)  is  the  chord  of  an  arc  of  60°,  and  therefore  equal 
to  the  radius.  (Trig.  95.)  The  chord  of  half  this  arc,  as 
BO,  is  the  side  of  a  polygon  of  12  equal  sides.  By  repeat- 
edly bisecting  the  arc,  and  finding  the  chord,  we  may  obtain 
the  side  of  a  polygon  of  an  immense  number  of  sides.  Or 
we  may  calculate  the  sine,  which  will  be  half  the  chord  of 
double  the  arc  ;  (Trig.  82,  cor.)  and  the  tangent,  which  will 
be  half  the  side  of  a  similar  circumscribed  polygon.  Thus 
the  sine  AP  (Fig.  7.)  i^  half  of  AB,  a  side  of  the  inscribed 
hexagon  ;  and  the  tangent  NO  is  half  of  NT,  a  side  of  the 
circumscribed  hexagon.  The  difference  between  the  sine 
and  the  arc  AG  is  less,  than  the  difference  between  the  sine 
and  the  tangent.  In  the  section  on  the  computation  of  the 
canon,  (Trig.  223.)  by  12  successive  bisections,  beginning 
with  60  degrees,  an  arc  is  obtained  which  is  the  ojjig  o(  the 
whole  circumference. 

Theco^neof  this,  if  radius  be  1,  is  found  to  be  .99999996732 
The  sine  is  .00025566346 

sine     _ 
And  the  tangent  =  — ^  (Trig.  228.)  =.00025566347 

costne 


The  diff  between  the  sine  and  tangent  is  only    .00000000001 
And  the  difference  between  the  sine  and  the  arc  is  still  less. 

*  In  this  Dcanner,  the  Supplement  to  Playfair'' s  Euclid  is  referred  loin  this 
work. 


It3         •  MENSURATION  OF 

Taking  then  .000255663465  for  the  length  of  the  arc,  mul- 
tiplying by  24567,  and  retaining  8  places  of  decimals,  we 
have  6.28318531  for  the  whole  circumference,  the  radius 
being  1.     Half  of  this, 

3.14159265 
is  the  circumference  of  a  circle  whose  radius  is  ^,  and  diam 
eter  1. 

22.  If  this  be  multiplied  by  7,  the  product  is  2l.99-|-  or 
22  nearly.     So  that, 

Diam  :  Circum:  ;7  !  22,  nearly. 

If  3.MI59265  be  multiplied  by  113,  the  product  is 
354.99994-,  or  355,  very  nearly.     So  that, 

Diam  :  Circum::  113  :  355,  very  nearly. 

The  first  of  these  ratios  was  demonstrated  by  Archimedes. 

There  are  various  methods,  principally  by  infinite  series 
and  fluxions,  by  which  the  labour  of  carrying  on  the  approx- 
imation to  the  periphery  of  a  circle  may  be  very  much 
abridged.  The  calculation  has  been  extended  to  nearly  150 
places  of  decimals.*  But  four  or  five  places  are  sufficient 
for  most  practical  purposes. 

After  detertnining  the  ratio  between  the  diameter  and  the 
circumference  of  a  circle,  the  following  problems  are  easily 
solved. 


rROBLEM    I. 

To  find  the  circumference  of  a  circle  from  its  diameter. 
Multiply  THE  diameter  by  3. 14l59.'f 

Multiply  (he  diameter  by  22  and  divide  the  product  by  7. 
Or,  multiply  the  diameter  by  355,  and  divide  the  product  by 
113.  (Art.  22.) 

Ex.  1.  If  the  diameter  of  the  earth  be  79:30  miles,  what 
is  the  circumference  i*  Ans.  249128  miles. 

2.  How  many  miles  does  the  earth  move,  in  revolving 
rodnd  the  sun  ;  supposing  the  orbit  to  bea  circle  whose  di- 
ameter is  190  million  miles  ?  Ans.  596,992,100. 

*  See  note  A. 

t  In  many  cases,  3.1416  will  be  sufficiently  accurate. 


THE  CIRCLE.  17 

What  is  the  circumference  of  a  circle  whose  diameter  is 
769843  rods  ? 


PROBLEM    II. 

To  find  the  diameter  of  a  circle  from  its  circumference, 

24.  Divide  the  circumference  by  3. 14 159. 

Multiply  the  circumference  by  7,  and  divide  the  product  by  22. 
Or,  multiply  the  circumference  by  113,  and  divide  the  pro- 
duct by  355.  (Art.  22.) 

Ex.  1.  If  the  circumference  of  the  sun  be  2,800,000 
miles,  what  is  his  diameter?  Ans.  891,267. 

2.  What  is  the  diameter  of  a  tree  which  is  5\  feet  round? 

25.  As  multiplication  is  more  easily  performed  than  division, 
there  will  be  an  advantage  in  exchanging  the  divisor  '3. 14159 
for  a  multiplier  which  will  give  the  same  result.  In  the  pro- 
portion 3.14159  :  l::rircum  :  Diam. 

to  find  the  fourth  term,  we  may  divide  the  second  by  the 
first,  and  multiply  the  quotient  into  the  third.  Now  l-~ 
3.14159=0.31831.  If  then  the  circumference  of  a  circle 
be  multiplied  by  .31831,  the  product  will  be  the  diameter.* 

Ex.  1.  If  the  circumference  of  the  moon  be  6850  mileSj 
what  is  her  diameter.'*  Ans.  2180. 

2.  If  the  whole  extent  of  the  orbit  of  Saturn  be  5650  mil- 
lion miles,  how  far  is  he  from  the  sun  ? 

3.  If  the  periphery  of  a  wheel  be  4  feet  7  inches,  what  is 
its  diameter. 

problem    III. 

To  find  the  length  of  an  arc  of  a  circle, 

26.  As  360°,  to  the  number  of  degrees  in  the. arc  ; 

So  is  the  circumference  of  the  circle,  to  the  length  of  the  arc. 

The  circumference  of  a  circle   being  divided  into  360°, 

(Trig.  73.)  it  is  evident  that  the  length  of  an  arc  of  any  less 

number  of  degrees  must  be  a  proporlional  part  of  the  whole. 


See  Note  B. 

4 


18  MENSURATION  OF 

Ex.  What  is  the  length  of  an  arc  of  16°,  in  a  circle  whose 
radius  is  50  feet  ? 

The  circumference  of  the  circle  is  314.159  feet.  (Art.  28.) 
Then  360  :  16::314.159  :  13.96  feet. 

2.  If  we  are  95  millions  of  miles  from  the  sun,  and  if  -he 
earth  revolves  round  it  in  365i  days,  how  far  are  we  carried 
in  24  hours  f  Ans.  i  million  634  thousand  mites. 

27.  The  length  of  an  arc  may  also  be  found,  by  multiply- 
ing the  diameter  into  the  number  of  degrees  in  the  arc.  and 
this  product  into  .0087266,  which  is  the  lei^gth  of  one  de- 
gree, in  a  circle  whose  diameter  is  1.  For  3.14159~.360= 
0.0087266.  And  in  different  circles,  the  circumferences,  and 
of  course  the  degrees,  are  as  the  diameters.  (Sup.  Euc.  8,  1.) 

Ex.  1.  What  is  the  length  of  an  arc  of  10°  15'  in  a  circle 
whose  radius  is  68  rods  .-^  Ans.  12.165  rods. 

2.  If  the  circumference  of  the  earth  be  24913  miles,  what 
is  the  length  of  a  degree  at  the  equator? 

28.  The  length  of  an  arc  is  frequently  required,  when  the 
number  of  degrees  is  not  given.  But  if  the  radius  of  the  cir- 
cle, and  either  the  chord  or  the  height  of  the  arc,  be  known  ; 
the  number  of  degrees  may  be  easily  found. 

Let  AB  (Fig.  9.;  be  the  chord,  and  PO  the  height,  of  the 
arc  AOB.  As  the  angles  at  P  are  right  angles,  and  AP  is 
equal  to  BP ;  (Art  18.  Def.  4.)  AO  is  equal  to  BO.  (Euc. 
4.  1.)     Then 

BP  is  the  sine,  CP  the  cosine.  ?     r  i  7r  u  a  /-kr> 

OP  the  versed  sine,  and  BO  the  chord  \  °*^^«iA  ^he  arc  AOB. 

And  in  the  right  angled  triangle  CBP, 

m     Vt       ^  BP  *  ^'n  BCP  or  BO 
^D  .  IV..  ^^p  ^  (^^gBCPoi  BO 

Ex.  1.  If  the  radius  CO  (Fig.  9.)  =25.  and  the  chord 
AB  =  43  3 ;  what  is  the  length  of  the  arc  AOB  ? 

CB  :  R:  :BP  :  Sin  BCP  or  BO=60°  very  nearly. 

The  circumference  of  the  circle  =3.14159x50=157.08. 
And  360°  :  60°::1 57.08  :  26. 1 8 =OB.Therefore  AOB  =52.36. 

2.  What  is  the  length  of  an  arc  whose  chord  is  216^  in  a 
circle  whose  radius  is  125?  Ans.  261.8. 


THE  CIRCLE.  19 

29.  If  only  the  chord  and  the  height  of  an  arc  be  given, 
the  radius  of  the  circie  may  be  found,  and  then  the  length  of 
the  arc. 

If  BA  (Fig  9.)  be  the  chord,  andPO  the  height  of  the  arc 
AOB,  then  (Euc.  35.3.) 

DP=-^  •     AndD0=0P+DP=0P+-^5p-  • 

That  is,  the  diameter  is  equal  to  the  height  of  the  arc,  -f- 
the  square  of  half  the  chord  divided  by  the  height. 

The  dianiL'ter  being  found,  the  length  of  the  arc  may  be 
calculated  by  the  two  preceding  articles. 

Ex.  1  If  the  chord  of  an  arc  be  173.2,  and  the  height  50, 
what  is  the  length  of  the  arc  1 

_  86^6' 

The  diameter  =50-|--^   =200.  The  arc  contains  120°; 

(Art.  28.)  and  its  length  is  209.44.     (Art.  26.) 

2.  What  is  the  length  of  an  arc  whose  chord  is  120,  and 
height  45.?  Ans.  160.8.* 


PROBLEM    IV. 

To  find  the  area  of  a  circle. 
30.  Multiply   the  square  of   the  diameter   by  the  decimals 

.7854. 

Or, 

Multiply  half  the  diameter  into  half  the  circum- 
ference. Or,  multiply  the  whole  diameter  into  the  whole 
circumference,  and  take  \  of  the  product. 

The  area  of  a  circle  is  equal  to  the  product  of  half  the 
diameter  into  half  the  circumference;  (Sup.  Euc.  5,  1.)  or 
which  is  the  same  thing,  {  the  product  of  the  diameter  and 
circumference.  If  the  diameter  be  1,  the  circum/erence  is 
3.14159;  (Art.  23.)  one  fourth  of  which  is  0.7S54  nearly. 
But  the  areas  of  different  circles  are  to  each  other,  as  the 
squares  of  their  diameters,  (Sup.  Euc  7,  l.f)  The  area  of 
any  circle,  therefore,  is  equal  to  the  product  of  the  square 

*>  See  note  C.  tEuclid2,  12. 


20  MENSURATION  OF 

of  its  diameter  into  0.7854,   which  is  the  area  of  a  circle 
whose  diameter  is  1. 

Ex.  1.  What  is  the  area  of  a  circle  whose  diameter  is  623 
feet?  Ans.  304836  square  feet. 

2.  How  many  acres  are  there  in  a  circular  island  whose 
diameter  is  124  rods  ?  Ans.  76  acres,  and  76  rods. 

3.  If  the  diameter  of  a  circle  he  113,  and  the  circumfer- 
ence .355,  what  is  the  area?  Ans.  10029. 

4.  How  many  square  yards  are  there  in  a  circle  whose 

diameter  is  7  feet  ? 

31.  If  the  circumference  of  a  circle  he  given,  the  area  may 
be  obtained,  by  first  finding  the  diameter ;  or,  without  finding 
the  diameter,  by  multiplying  the  square  of  the  circumference 
by  .07958. 

For,  if  the  circumference  of  a  circle  be  1,  the  diameter 
=  1-7-3.14159=0.31831  ;  and  i  the  product  of  this  into  the 
circumference  is  .07956  the  area.  But  the  areas  of  differ- 
ent circles,  being  as  the  squares  of  iheir  diameters,  are  also 
as  the  squares  of  their  circumferences.  (Sup.  Euc.  8.  1.) 

Ex.  1-  If  the  circumference  of  a  circle  be  136  feet,  what 
is  the  area  ?  Ans.  1472  feet. 

2.  What  is  the  surface  of  a  circular  fish-pond,  which  is  10 
rods  in  circumference .'' 

32.  If  the  area  of  a  circle  he  given,  the  diameter  may  be 
found,  by  dividing  the  area  by  .7854,  and  extracting  the 
square  root  of  the  quotient. 

This  is  reversing  the  rule  in  art.  30. 

Ex.  1.  What  is  vhe  diameter  of  a  circle   whose  area  is 

380.1336  feet.^Ans.380.1336~.7854=484.AndV484=22. 

2.  What  is  the  diameter  of  a  circle  whose  area  is  19.635.^ 

33.  The  area  of  a  circle,  is  to  the  area  of  the  circum- 
scribed  square  ;  as  .7854  to  1  ;  and  to  that  of  the  inscribed 
square  as  .7854  to  \. 

Let  ABDF  (Fig.  10.)  be  the  inscribed  square  and  LMNO 
the  circumscribed  square,  of  the  circle  ABDF.  The  area 
of  the  circle  is  equal  to  AD  ^  X.7854.  (Art.  30)  Butthearea 
of  the  circumscribed  square  (Art.  4.)  is  equal  to  0N^=  ^D*. 
And  the  smaller  square  is  half  of  the  larger  one.     For  the 


THE  CIRCLE.  ^1 

laiter  contains  8  equal  triangles,  of  which  the  former  contains 
only  4. 

Ex.  VVhat  is  the  area  of  a  square  inscribed   in  a   circle 
whose  area  is  159.?  Ans.  .7854  :  -^::159  :  101.22. 


PROBLKAI    V. 

To  find  the  area  of  a  sector  of  a  circle. 

34.  Multiply  the  radius  into  half  the  length  of  the  arc. 

Or, 

As    360,  TO    THE    NUMBER    OF    DEGREES  IN    THE    ARC  ; 

So    IS  THE    AREA    OF    THE    CIRCLE,  TO    THE    AREA  OF  THE  SECTOR. 

It  is  evident,  that  the  area  of  the  sector  has  the  same  ratio 
to  the  area  of  the  circle,  which  the  length  of  the  arc  has  to 
the  length  of  the  whole  circumference  ;  or  which  the  number 
q{  degrees  m  the  arc  has  to  the  number  of  degrees  in  the  cir- 
cumference. 

Ex.  1.  If  the  arc  AOB  (Fig.  9.)  be  120^  and  the  diam- 
eter of  the  circle  226  ;  what  is  the  area  of  the  sector  AOBC? 
The  area  of  the  whole  circle  is  401 15.  (Art.  30.) 

And  360=^  :  120°: : 401 15  :  1337 If,  the  area  of  tiio  sector. 

2.  What  is  the  area  of  a  quadrant  whose  radius  is  621  .'' 

3.  What  is  the  area  of  a  semi-circle  whose  diameter  is  328  ^ 

4.  What  is  the  area  of  a  sector  which  is  less  than  a  semi- 
circle, if  the  radius  be  15,  and  the  chord  of  its  arc  12  .? 

Half  the  chord  is  the  sine  of  23°  34'f  nearly.     (Art.  23.) 

The  whole  arc,  tl'.en.  is         47°     9'i 

The  area  of  the  circle  is         706  86 

And  360°  :  47°  9'^: :  706.86  :  92.6  the  area  of  the  sector. 

5.  If  the  arc  ADB  (Fig.  9.)  be  240  degrees,  and  the  radiuj^ 
of  the  circle  113,  what  is  the  area  of  the  sector  ADBC  ^ 


PROBLEM    VI. 

To  find  the  area  of  a  segment  of  a  circle, 
35.  Find   the  area  of  the  sector  which  has  the  same  arc, 

AND  also  THE  AREA  OF  THE  TRIANGLE  FORMED  BV  THE  CHORD  OF 
THE  SEGMENT  AND  THE  RADH  OF  THE  SECTOR. 


«2  MENSURATION  OF 

Then,  if  the  segment  be  less  than  a  semi-circle,  subtract 
the  area  of  thf.  triangle  from  the  area  of  the  sector.  but, 
if  the  segment  be  greater  than  a  semi-circle,  add  th«<:  area 
of  the  triangle  to  the  area  of  the  sector. 

If  the  triangle  ABC  (Fig.  9.)  be  taken  from  the  sector 
AOBC,  it  IS  evident  the  difference  will  be  the  segment  AOBP, 
less  than  a  semi-circle.  And  if  the  same  tri-dnii;le  be  added 
to  the  sector  ADBC,  the  sum  will  be  liie  segment  ADBP, 
greater  than  a  semi-circle. 

The  area  of  the  triangle  (Art  8.)  is  equal  to  the  product 
of  half  the  chord  AB  into  CP  which  is  the  difference  be- 
tween the  radius  and  PO  the  height  of  the  segment.  Or 
CP  is  the  cosine  of  half  the  arc  BOA.  If  this  cosine,  and 
the  chord  of  the  segment  are  not  given,  they  may  be  found 
from  the  arc  and  the  radius. 

Ex.  1.  If  the  arc  AOB  (Fig.  9.)  be  120°,  and  the  radius 
of  the  circle  be  113  feet,  what  is  the  area  of  the  segment 
AOBP  ? 

In  the  right  angled  triangle  BCP, 
R  :  BC::Sin  BCO  :  BP=r.97.86,  half  the  chord.  (Art  28.) 

The  cosine  PC  =  iCO  (Trig.  96,  Cor.)  =56  5 

The  area  of  the  sector  AOBC  ^Art.  34.)  =    1 337  J. 67 

The  area  of  the  triangle  ABC  =  BP  xPC        =     £528.97 


The  area  of  the  segment,  therefore,  =     7842.7 

2,  If  the  base  of  a  segment,  less  than  a  semi-circle,  be  10 
feet,  and  the  radius  of  the  circle  12  feet,  what  is  the  area  of 
the  segment  ? 

The  arc  of  the  segment  contains     49]  degrees.  (Art.  28.) 
The  area  of  the  sector  =6 1 .89  (Art.  34.) 

The  area  of  the  triansie  =54.54 


And  the  area  of  the  segment       =  7.35  square  feet. 

3.  What  is  the  area  of  a  circular  segQient,  whose  height  is 
19.2  and  base  10?  Ans.  947.86. 


THE  CIRCLE.  23 

4.  What  is  the  area  of  the  segment  ADBP,  (Fig.  9.)  if  the 
base  AB  be  195.7,  and  the  height  PD  169.5  ? 

Ans.  32272.* 

36.  The  area  of  any  figure  which  is  bounded  partly  by 
arcs  of  circles,  and  partly  by  right  hnes,  may  be  calculated, 
by  rinding  the  areas  of  the  segments  under  the  arcs,  and  then 
the  nrea  of  the  rectilinear  space  between  the  chords  of  the 
arcs  and  the  other  right  lines. 

Tims  the  Gothic  arch  ACB,  (Fig.  11.)  contains  the  two 
segments  ACH,  BCD,  and  the  plane  triangle  ABC. 

Ex  If  AB  (Fig.  11.)  be  110,  each  of  the  lines  AC  and 
B(  00.  and  the  iieight  of  each  of  the  segments  ACH,  BCD 
10.435  ;  what  is  the  area  of  the  whole  figure  f 

The  areas  of  the  two  sefijments  are  1404 

The  area  of  the  triangle  ABC  is  4593.4 


And  the  whole  figure  is  5997.4 


FROBLEM    Vn. 


To  find  the  area  nf  a  circular  zone. 
37.  From  the  area  of  the    whole  circle,  subtract  the  two 


SEGMENTS    ON    THE    SIDES    OF    THE    ZONE. 


If  from  the  whole  circle  (Fig.  12.)  there  be  taken  the  two 
segments  ABC  and  DFH,  there  will  remain  the  zone 
ACDH. 

Or,  the  area  of  the  zone  may  be  found,  by  subtracting  the 
segment  ABC  from  the  segment  HBD :  Or,  by  adding  the 
two  small  segments  GAH  and  VDC  to  the  trapezoid  ACDH. 
See  art.  36. 

The  latter  method  is  rather  the  most  expeditious  in  prac- 
tice, as  the  two  segments  at  the  end  of  the  zone  are  equal, 

Ex.  1.  ^Vhat  is  the  area  of  the  zone  ACDH,  (Fig.  12.)  if 
AC  is  7.7«,  DH  6.93,  and  the  diameter  of  the  circle  8  ^ 

*  For  the  method  of  finding  the  areas  of  segments  by  a  tahle^  ?ee  note  P 


24  MENSURATION  OF 

The  area  ofthewln/io  circle  is  50.26 

of  the  segment  ABC         17.32 
of  the  segment  DFH  9.82 

of  the  zone  AC l)H  23.12 

2.  What  is  the  area  of  a  zone,  one  side  of  which  is  23.25, 
and  the  other  side  20.8,  in  a  circle  whose  diameter  is  24  ? 

Ans.  208. 

38  If  the  diameter  of  the  circle  is  not  given,  it  may  be 
found  from  the  sides  and  the  breadth  of  the  zone. 

Let  the  centre  of  the  circle  be  at  O.  (Fig.  12.)  Draw 
ON  perpend'cular  to  AH,  NM  perpendicular  to  LR,  and 
HP  perpenJicular  to  AL.     Then 

AN=iAH,  (Euc.  3.  3.)        MN=i(LA4-RH) 
LM=iLK,  (Euc.  2.  6.)         PA=:LA-RH. 

The  triangles  APH  and  OMN  are  similar,  because  the 
sides  of  one  are  perpendicular  to  tliose  of  the  other,  each  to 
each.     Therefore 

PH:PA::MN:MO 

MO  being  found,  we  have  ML-MO=OL. 

And  the  radius  CO^V'OL^+CL^  (Euc.  47.  1.) 

Ex.  If  the  breadth  of  the  zone  ACDH  (Fig.  12.)  be  6.4, 
$nd  the  sides  6.8  and  6  ;  what  is  the  radius  of  the  circle  ? 
PA=3.4-3=0.4.         And  MN  =  i(3.4-f  o)  =  3.2. 
Then  6.4  :  0.4::3.2  :  0  2=M0.     And  3.2-0.2=3=OL 
And  the  radius  C0='/3"^ +(3.4)3  =4.534. 


PROBLEM    VIII. 

To  find  the  area  of  a  lhne  or  crescent, 

39.  Find  the  difference  of  the  two  segments  which  are  be- 
tween THE  ARCS   OF    THE    CRESCENT    AND  ITS    CHORD. 

If  the  segment  ABC   (Fig   14.)  be  taken   from  the   seg- 
ment ABD;  there  will  remain  the  lune  or  crescent  ACBD. 

Ex.  If  the  chord  AB  be  88,  the  height  CH  20,   and  the 
height  DH  40 ;  what  is  the  area  of  the  crescent  ACBD  ? 


THE  CIRCLE.  25 

The  area  of  the  segment  ABD  is  2698 

of  the  segment  ABC  1220 

of  the  crescent  ACBD  1478 


PROBLEM    IX. 

To  find  the  area  of  a  ring,  included  between  the  peripheries 
of  two  concentric  circles. 

40.  Find  the  difference  of  the  areas  of  the  two  circles. 

Or, 

Muhiply  the  product  of  the  sum  and  difference  of  the  two 
diameters  by  .7854. 

The  area  of  the  ring  (Fig.  13.)  is  evidently  equal  to  the 
difference  between  the  areas  of  the  two  circles  AB  and  A  B'. 

But  the  area  of  each  circle  is  equal  to  the  square  of  its 
diameter  multiplied  into  .7854.  (Art.  o9.)  And  the  differ- 
ence of  these  squares  is  equal  to  the  product  of  the  sum  and 
difference  of  the  diameters.  (Alg.  235.)  Therefore  the 
area  of  the  ring  is  equal  to  the  product  of  the  sum  and  differ- 
ence of  the  two  diameters  multiplied  by  .7854. 

Ex.  1.  If  AB  (Fig.  13.)  be  221,  and  A'B'  106,  what  is  the 
area  of  the  ring  .** 

Ans.  ("22^  X. 7854) -( foe 'x. 7854)  =  29535. 

2.  If  the  diameters  of  Saturn's  larger  ring  be  205,000  and 
190.000  miles,  how  many  square  miles  are  there  on  one  side 
of  the  ring  ^ 

Ans.  395000  X  1 5000  X  .7854=4,653,495,000. 


PROMISCUOUS  EXAMPLES  OF  AREAS. 

Ex.   1.  What  is   the  expense  of  paving  a  street  20  rods 
long,  and  2  rods  wide,  at  5  cents  for  a  square  foot  ^ 

Ans.  544^  dollars. 


26  MENSURATION  OF 

2.  If  an  equilateral  triangle  contains  as  many  square  feet 
as  there  are  inches  in  one  of  its  sides  ;  what  is  the  area  of  the 
triangle  ? 

Let   x=  the  number  of  square  feet  in  the  area. 

X 

Then  7^=  the  number  of  linear  feet  in  one  of  the  sides. 

476 
Reducing  the  equation,  a?= -^  =  332.55  the  area. 

3.  What  is  the  side  of  a  square  whose  area  is  equal  to 
that  of  a  circle  452  feet  in  diameter  ? 

Ans.  \/(452)2  X. 7854=400.574.    (Art.  30  and  7.) 

4.  What  is  the  diameter  of  a  circle  which  is  equal  to  a 
square  whose  side  is  36  feet  ? 

Ans.  v/(36)2-i-0.7854=40.6217.  (Art.  4.  and  32.) 

5.  What  is  the  area  of  a  square  inscribed  in  a  circle  whose 
diameter  is  132  feet.^ 

Ans.  8712  square  feet.    (Art.  33.) 

6.  How  much  carpeting,  a  yard  wide,  will  be  necessary  to 
cover  the  floor  of  a  room  which  is  a  regular  octagon,  the  sides 
being  8  feet  each  f  Ans.  34|  yards. 

7.  If  the  diagonal  of  a  square  be  16  feet,  what  is  the  area  ? 

Ans.  128  feet.  (Art.  14.) 

8.  If  a  carriage  wheel  four  feet  in  diameter  revolve  300 
times,  in  going  round  a  circular  green  ;  what  is  the  area  of  the 
green  f 

Ans.  41541  sq.  rods,  or  25  acres,  3  qrs.  and  34^  rods. 

9.  What  will  be  the  expense  of  papering  the  sides  of  a 
room,  at  10  cents  a  square  yard  ;  if  the  room  be  21  feet  long, 
18  feet  broad,  and  12  feet  high  ;  and  if  there  be  deducted  3 
windows,  each  5  feet  by  3,  two  doors  B  feet  by  41,  and  one 
fire-place  6  feet  by  4i  ?  Ans.  8  dollars  80  cents. 

10.  If  ^  circular  pond  of  water  10  rods  in  diameter  be  sur- 
rounded by  a  gravelled  walk  8}  feet  wide  ;  what  is  thci  area  of 
the  walk  ?  Ans.  16i  sq.  rods.  (Art.  40.) 


SUPERFICIES.  27 

11.  If  CD  (Fig.  17,)  the  base  of  the  isosceles  triangle 
VCD,  be  60  feet,  and  the  area  1200  feel ;  and  if  there  be  cut 
off,  by  the  line  LG  parallel  to  CD,  the  triangle  VLG,  whose 
area  is  432  feet ;  what  are  the  sides  of  the  latter  triangle  f 

Ans.  30,  30,  and  36  feet. 

12.  What  is  the  area  of  an  equilateral  triangle  inscribed  in 
a  circle  whose  diameter  is  52  feet  ? 

Ans.  878.15  sq.  feet. 

13.  If  a  circular  piece  of  land  is  enclosed  by  a  fence,  in 
which  10  rails  make  a  rod  in  length ;  and  if  the  field  contains 
as  many  square  rods,  as  there  are  rails  in  the  fence;  what  is 
the  value  of  the  land  at  120  dollars  an  acre  ? 

Ans.  942  48  dollars. 

14.  If  the  area  of  the  equilateral  triangle  ABD  (Fig.  9.) 
be  219.5375  feet ;  what  is  the  area  of  the  circle  OBDA,  in 
which  the  trian^jle  is  inscribed  .'* 

The  sides  of  the  triangle  are  each  22.5167.  (Art.  11.) 
And  the  area  of  the  circle  is  530.93 

15.  If  G  concentric  circles  are  so  drawn,  that  the  space 
between  the  least  or  1st,  and  the  2d  is  21.2058, 

between  the  2d  and  3d  35. 343, 

between  the  3d  and  4th  49.4802, 

between  the  4lh  and  5th  63  6174, 

between  the  oth  and  6th  77.7546; 

what  are  the  several  diameters,  supposing  the  longest  to  be 

equal  to  6  times  the  shortest.^ 

Ans.  3,6,  9,  12,15,  and  18. 


SECTION   III. 


SOLIDS  BOUNDED  BY  PLANE  SURFACES. 


^,       T^  ,4    PRISM  is  a  solid  bounded 

Art.  41.     Definition  L  A  ^^  pj^„^  ^^^^^^  ^^  ^^^^3^  ^^^ 

of  which  are  parallel,  similar,  and   equal;  and,the  others  are 
parallelograms. 

II.  The  parallel  planes  are  sometimes  called  the  bases  or 
ends  ;  and  the  other  figures,  the  sides  of  the  prism.  The  lat- 
ter taken  together  constitute  the  lateral  surface. 

III.  A  prism  is  right  or  oblique^  according  as  the  sides  are 
perdendicular  or  oblique  to  the  bases. 

IV.  The  height  of  a  prism  is  the  perpendicular  distance 
between  the  planes  of  the  bases.  In  a  right  prism,  there- 
fore, the  height  is  equal  to  the  length  of  one  of  the  sides. 

V.  A  Farallelopiptd  is  a  prism  whose  bases  are  parallelo- 
grams, 

VI.  A  Cube  is  a  solid  bounded  by  six  equal  squares.  Jt 
is  a  ric^ht  prism  whose  sides  and  bases  are  all  equal. 

VII.  A  Pyramid  is  a  solid  bounded  by  a  plane  figure  call- 
ed the  base,  and  several  triangular  planes,  proceeding  from 
the  sides  of  the  base,  and  all  terminating  in  a  single  point. 
These  triangles  taken  together  constitute  the  lateral  surface. 

VIII.  A  pyramid  is  regular,  if  its  base  is  a  regular  poly- 
gon.  and  if  a  line  from  the  centre  of  the  base  to  the  vertex  of 
the  pyramid  is  perpendicular  to  the  base.  This  line  is  called 
the  axis  of  the  pyramid. 

IX.  The  height  of  a  pyramid  is  the  perpendicular  distance 
from  the  summit  to  the  plane  of  the  base.  In  a  regular  pyra- 
mid, it  is  the  length  of  the  axis. 

X.  The  slont-height  of  a  regular  pyramid,  is  the  distance 
from  the  summit  to  the  middle  of  one  of  the  sides  of  the  base. 

XL  A  frustum  or  trunk  of  a  pyramid  is  a  portion  of  the 
solid  next  the  base,  cut  off  by  a  plane  parallel  to  the  base. 
The  height  of  the  frustum  is  the  perpendicular  distance  of 
tbe  two  parallel  planes.     The  slant-height  of  a  frustum  of  a 


MENSURATION  OF  SOLIDS.  29 

regular  pyramid,  is  the  distance  from  the  middle  of  one  of 
the  sides  of  the  base,  to  the  middle  of  the  corresponding  side 
in  the  plane  above.  It  is  a  line  passing  on  the  surface  of  the 
frustum,  through  the  middle  of  one  of  its  sides. 

XII.  A  Wedge  is  a  solid  of  five  sides,  viz.  a  rectangular 
base,  two  rhomboidal  sides  meeting  in  an  edge,  and  two  tri- 
angular ends  ;  as  ABHG.  (Fig  20.)  The  base  is  ABCD, 
the  sides  ate  ABHG  and  DCHG,  meeting  in  the  edge  GH, 
and  the  ends  are  BCH  and  ADG.  The  height  of  the  wedge 
is  a  perpendicular  drawn  from  any  point  in  the  edge,  to  the 
plane  of  the  base,  as  GP. 

XIII.  A  Prismoid  is  a  solid  whose  ends  or  bases  are  par- 
allel, but  not  similar,  and  whose  sides  are  quadrilateral.  It 
differs  from  a  prism  or  a  frustum  of  a  pyramid,  in  having  its 
ends  dissimilar.  It  is  a  rectaiigular  prismoid,  when  its  ends 
are  right  parallelograms. 

XIV.  A  linear  side  or  edge  of  a  solid  is  the  line  of  inter- 
section of  two  of  the  planes  which  form  the  surface. 

42.  The  common  measuring  vnit  of  solids  is  a  cuhe^  whose 
sides  are  squares  of  the  same  name.  The  sides  of  a  cubic 
inch  are  square  inches ;  of  a  cubic  foot,  square  feet,  &c. 
Finding  the  capacity,  solidity  *  or  solid  contents  of  a  body,  is 
finding  the  number  of  cubic  measures,  of  some  given  denom- 
ination contained  in  the  body. 

In  solid  measure. 

1728     cubic  inches  =1   cubic  foot, 
27     cubic  feel       =1   cubic  yard, 
4492|  cubic  feet       =1  cubic  rod, 
32768000     cubic  rods      =1  cubic  mile, 
282     cubic  inches  =1  ale  gallon, 
231     cubic  inches  =1   wine  gallon, 
21i0.42     cubic  inches  =1   bushel, 

1     cubic  foot   of   pure  water  weighs    1000 
Avoirdupois  ounces,  or  62^  pounds. 

*  See  note  E. 


3Q  MENSURATION  OF 

PROBLEM    I. 

To  find  the  solidity  of  a  prism. 

43.  Multiply  the  area  of  the  base  by  the  height. 

This  is  a  general  rule,  applicable  to  parallelepipeds  whether 
right  or  nblique,  cubes,  triangular  prisms,  (fee. 

As  surfaces  are  measured,  by  comparing  them  with  a  right 
paraVelogram  (Art.  3.) ;  so  solids  are  measured,  by  compar- 
ing them  with  a  r'l^hi par allelopiped. 

If  ABCD  (Fig.  1.)  be  the  base  of  a  right  parallelopiped, 
as  a  stick  of  timber  standing  erect,  it  is  evident  that  the 
number  of  cubic  feet  contained  in  one  foot  of  the  height,  is 
eq-.ial  to  the  number  of  square  feet  in  the  area  of  the  base. 
And  if  the  solid  be  of  any  other  height,  instead  of  one  foot, 
the  contents  must  have  the  same  ratio.  For  parallelopipeds 
of  the  same  base  are  to  each  other  as  their  heights.  (Sup. 
Euc.  9.  3.)  The  solidity  of  a  right  parallelopiped,  there- 
fore, is  equal  to  the  product  of  its  length,  breadth^  and  thick- 
ness.    See  Alg.  523. 

And  an  oblique  parallelopiped  being  equal  to  a  right  one  of 
ihe  same  base  and  altitude,  (Sup.  Euc.  7,  3.)  is  equal  to  the 
area  of  the  base  multiplied  into  the  perpendicular  height. 
This  is  true  also  of  prisms,  whatever  be  the  form  of  their 
bases.  (Sup.  Euc.  2.  Cor.  to  8.  3.) 

44.  As  the  sides  of  a  cube  are  all  equal,  the  solidity  is 
found  by  cubing  one  of  its  edges.  On  the  other  hand,  if  the 
solid  contents  be  given,  the  length  of  the  edges  may  be  found, 
by  extracting  the  cube  root, 

45.  When  solid  measure  is  cast  by  Duodecimals,  it  is  to  be 
observed  that  inches  are  not  primes  of  feet,  but  thirds.  If 
the  unit  is  a  cubic  foot,  a  solid  which  is  an  inch  thick  and  a 
foot  square  is  a  prime  ;  a  parallelopiped  a  foot  long,  an  inch 
broad,  and  an  inch  thick  is  a  second,  or  the  twelfth  part  of  a 
prime ;  and  a  cubic  inch  is  a  third,  or  a  twelfth  part  of  a 
second.  A  linear  inch  is  ^3  of  a  foot,  a  square  inch  yi^  of 
a  foot,  and  a  cubic  inch  jj\^  of  a  foot, 

Ex.  1.  What  are  the  solid  contents  of  a  stick  of  timber 
which  is  31  feet  long,  1  foot  3  inches  broad,  and  9  inches 
thick  ?  Ans.  29  feet  9",  or  29  feet  108  inches. 


SOLIDS.  31 

2.  What  is  the  solidity  of  a  wall  which  is  22  feet  long,  12 
feet  high,  and  2  feet  6  inches  thick  f 

Ans.  660  cubic  feet, 

3.  What  is  the  capacity  of  a  cubical  vessel  which  is  2  feet 

3  inches  deep  f 

Ans.  llF.  4'  8"  3'",  or  11  feet  675  inches. 

4.  If  the  base  of  a  prism  be  108  square  inches,  and  the 
height  36  feet,  what  are  the  solid  contents  ? 

Ans.  27  cubic  feet. 

5.  If  the  height  of  a  square  prism  be  2\  feet,  and  each 
side  of  the  base  10^  feet,  what  is  the  solidity  f 

The  area  of  the  base  =10i  X  lO^X  106|  sq.  feet. 
And  the  solid  contents  =106|  X2i  =  240i  cubic  feet. 

6.  If  the  height  of  a  prism  be  23  feet,  and  its  base  a  regu- 
lar pentagon,  whose  perimeter  is  18  feet,  what  is  the  soliiity  ? 

Ans.  512.84  cubic  feet. 

46.  The  number  o( gallons  or  bushels  which  a  vessel  will 
contain  may  be  found,  by  calculating  the  capacity  in  inches^ 
and  then  dividing  by  the  number  of  inches  in  1  gallon  or 
bushel. 

The  weight  of  water  in  a  vessel  of  given  ditnensions  is 
easily  calculated ;  as  it  is  found  by  experiment,  that  a  cubic 
foot  of  pure  water  weighs  1000  ounces  Avoirdupois.  For 
the  weight  in  ounces,  then,  multiply  the  cubic  feet  by  1000  5 
or  for  the  weight  in  pounds,  multiply  by  62i. 

Ex.  1.  How  many  ale  gallons  are  there  in  a  cistern  which 
is  1 1  feet  9  inches  deep,  and  whose  base  is  4  feet  2  inches 
square/ 

The  cistern  contains  352500  cubic  inches  ; 

And  352500—282=1259. 

2.  How  many  wine  gallons  will  fill  a  ditch  3  feet  11  inches 
wide,  3  feet  deep,  and  462  feet  long  ?  Ans.  4060S. 

3.  What  weight  of  water  can  be  put  into  a  cubical  vessel 

4  feet  deep  ?  Ans.  4000  Ib^. 


a^  MENSURATION  OF 

PROBLEM    II. 
To^ndMc  LATERAL  SURFACE  of  a  RIGHT  PRISM. 

47.  Multiply  the  length  into  the  perimeter  of  the  base. 
Each  of  the   sides  of  the  prism  is  a  right  parallelogram, 

whose  area  is  the  product  of  its  length  and  breadth.  But 
the  breadth  is  one  side  of  the  base  ;  and  therefore,  the  sum 
of  the  breadths  is  equal  to  the  perimeter  of  the  base, 

Ex.  1.  If  the  base  of  a  right  prism  be  a  regular  hexagon 
whose  sides  are  each  2  feet  3  inches,  and  if  the  height  be  16 
feet,  what  is  the  lateral  surface  ^ 

Ans.  216  square  feet. 

If  the  areas  of  the  two  ends  be  added  to  the  lateral  sur- 
face, the  sum  will  be  the  whole  surface  of  the  prism.  And 
the  superficies  of  any  solid  bounded  by  planes,  is  evidently 
equal  to  the  areas  of  all  its  sides. 

Ex.  2.  If  the  base  of  a  prism  be  an  equilateral  triangle 
whose  perimeter  is  6  feet,  and  if  the  height  be  17  feet,  what  ^s 
the  surface  .^ 

The  area  of  the  triangle  is  1.732.  (Art.  11.) 
And  the  whole  surface  is  105.464. 

PROBLEM    III. 

To  find  the  solidity  of  a  pyramid. 

48.  Multiply  the  area  of  the  base  into  ^  of  the  height. 
The  solidity  of  a  prism  is  equal  to  the  product  of  the  area 

of  the  base  into  the  height.  (Art.  43.)  And  a  pyramid  is  ^ 
of  a  prism  of  the  same  base  and  altitude.  (Sup.  Euc.  15.  3. 
Cor.  1.)  Therefore  the  solidity  of  a  pyramid  whether  right 
or  oblique,  is  equal  to  the  product  of  the  base  into  \  of  the 
perpendicular  height. 

Ex.  1.  What  is  the  solidity  of  a  triangular  pyramid,  whose 
height  is  60,  and  each  side  of  whose  base  is  4  .'* 

The  area  of  the  base  is  6.928 

And  the  solidity  is  138.56. 

2.  Let  ABC  (Fig.  16.)  be  one  side  of  an  oblique  pyramid 
whose  base  is  6  feet  square;  letBC  be  20  feet,  and  make  an 
angle  of  70  degrees  with  the  plane  of  the  base ;  and  let  CP 
be  perpendicular  to  this  plane.  What  is  the  solidity  of  the 
pyramid  .'* 


SOLIDS.  33 

la  the  right  angled  triangle  BCP,  {Trig.  134.) 
R  :  BC.lSinB  :  PC  =  18.79. 

And  the  sohdity  of  the  pyramid  is  225.48  feet. 
3.  What  is  the  solidity  of  a  pyramid  whose  perpendicular 
height  is  72,  and  the  sides  of  whose  base  are  67,  54,  and  40  f 

Ans.  25920. 

PROBLEM  IV. 

To  find  the  lateral  surface  of  a  regular  pyramid. 
49.  Multiply  half  the  slant-height    into  the  perimeter  of 

THE    BASE. 

Let  the  triangle  ABC  (Fig.  18.)  be  one  of  the  sides  of  a 
regular  pyramid.  As  the  sides  AC  and  BC  are  equal,  the 
angles  A  and  B  are  equal.  Therefore  a  line  drawn  from  the 
vertex  C  to  the  middle  of  AB  \s perpendicular  to  AB.  The 
area  of  the  triangle  is  equal  to  the  product  of  half  this  per- 
pendicular into  AB.  (Art.  8.)  The  perimeter  of  the  base 
is  the  sum  of  its  sides,  each  of  which  is  equal  to  AB.  And 
the  areas  of  all  the  equal  triangles  which  constitute  the  lateral 
surface  of  the  pyramid,  are  together  equal  to  the  product  of 
the  perimeter  into  half  the  slant-height  CP. 

The  slant-height  is  the  hypothenuse  of  a  right  angled  tri- 
angle, whose  legs  are  the  axis  of  the  pyramid,  and  the  dis- 
tance from  the  centre  of  the  base  to  the  middle  of  one  of  the 
sides.  See  Def.  10. 

Ex.  1.  What  is  the  lateral  surface  of  a  regular  hexagonal 
pyramid,  whose  axis  is  20  feet,  and  the  sides  of  whose  base 
are  each  8  feet  f 

The  square  of  the  distance  from  the  centre  of  the  base  to 
one  of  the  sides  (Art.  16.)     =48. 

The  slant-height  (Euc.  47.  1.)      =\/4S-f- 20^  =21.16. 
And  the  lateral  surface  =21.16  X 4x6  =  507.84  sq.  {qqU 
2.  What  is  the  whole  surface  of  a  regular  triangular  pyra- 
mid whose  axis  is  8,  and  the  sides  of  whose  base  are  each 
20.78 .? 

The  lateral  surface  is  312 

The  area  of  the  base  is  187 

And  the  whole  surface!^'  499 


M  MENSURATION  OF 

3.  What  is  the  lateral  surface  of  a  regular  pyramid  whose 
axis  is  12  feet,  and  whose  base  is  18  feet  square  ? 

Ans.  540  square  feet. 

The  lateral  surface  of  an  oblique  pyramid  may  be  found, 
by  taking  the  sum  of  the  areas  of  the  unequal  triangles  which 
form  its  sides.^ 


PROBLEM    V. 

To  find  the  solidity  of  a  frustum  of  a  pyramid, 

50.  Add  together  the  areas  of  the  two  ends,  and  the  square 
root  of  the  product  of  these  areas  ;  and  multiply  tfi»  sum  by 
a  of  the  perpendicular  height  of  the  soltd. 

Let  CDGL  (Fig.  17.)  be  a  vertical  section,  through  the 
middle  of  a  frustum  of  a  right  pyramid  CDV  whose  base  is  a 
square, 

LetCD=<i,  LG=J,  RN=^k 

By  similar  triangles,  LG  :  CD:  :RV  :  NV. 

Subtracting  the  antecedents,  (Alg.  389.) 
LG  :  CD-LG::RV  :  NV~RV=RN. 
RNxLG      hb 


Therefore  RV: 


CD~LG~a-6 


The  square  of  CD  is  the  base  of  the  pyramid  CDV ; 

And  the  square  of  LG  is  the  base  of  the  small  pyramid  LGV. 

Therefore,  the  solidity  of  the  larger  pyramid  (Art,  48.)  is 

,  /         hb  \        ha^ 

CD-  X  i(RN+RV)=«=  X  i  (a+— J  =3^336 

And  the  solidity  of  the  smaller  pyramid  is  equal  to 

LG'xiRV=6=X3^j=3^. 

If  the  smaller  pyramid  be  taken  from  the  larger,  there  will 
remain  the  frustum  CDLG,  whose  solidity  is  equal  to 
ha^—hb^  a^-b^ 

Or,  because  "^a^  =ab,  (Alg.  259.) 


SOLIDS.  36 

Here  A,  the  height  of  the  frustum,  is  multiplied  into  a'  and 

6^,  the  areas  of  the  two  ends,  and  into  y/a^h*    the   square 
root  of  the  products  of  these  areas. 

In  this  demonstration,  the  pyramid  is  supposed  to  be 
square.  But  the  rule  is  equally  applicable  to  a  pyramid  of 
any  other  form.  For  the  solid  contents  of  pyrainids  are 
equal,  when  they  have  equal  heights  and  bases,  whatever  be 
the ^^urc  of  their  bases.  (Sup.  Euc.  14.  3.)  And  the  sec- 
tions parallel  to  the  bases,  and  at  equal  distances,  are  equal  to 
one  another.  (Sup.  Euc.  12.  3.  Cor.  2.)* 

Ex.  1.  If  one  end  of  the  frustum  of  a  pyramid  be  9  feet 
square,  the  other  end  6  feet  square,  and  the  height  36  feet, 
what  is  the  solidity  ^ 

The  areas  of  the  two  ends  are  81  and  36. 
The  square  root  of  their  product  is  54. 
And  the  solidity  of  the  frustum  =(81 4-36-}-54)X  12=2052, 

2.  If  the  height  of  'a  frustum  of  a  pyramid  be  24,  and  the 
areas  of  the  two  ends  441  and  12]  ;  what  is  the  solidity  I 

Ans.  6344. 

3.  If  the  height  of  a  frustum  of  a  hexagonal  pyramid  be 
48,  each  side  of  one  end  26,  and  each  side  of  the  other  end 
16;  what  is  the  solidity  .'*  Ans.  560o4. 


PROBLEM    ri. 

To  find  the  lateral  surface  of  a  frustvsi  of  a  regular 
pyramid, 

51.  Multiply  half  the  slant-height  bv  the  sum  of  the  perim- 
eters OF  the  two  ends. 

Each  side  of  a  frustum  of  a  regular  pyramid  is  a  trapezoid, 
as  ABCD.  (Fig.  19.)  The  slant-hei^/u  HP,  (Def.  11.) 
though  it  is  oblique  to  the  base  of  the  solid,  is  perpendicular 
to  the  line  AB.  The  area  of  the  trapezoid  is  equal  to  the 
product  of  half  this  perpendicular  into  the  sum  of  the  par- 
allel sides  AB  and  DC.  (\rf.  12)  Therefore  the  area  of 
all   the  equal   trapezoids  which  form  the  lateral  surface  of 

*  See  not&F. 


36  MENSURATION  OF 

the  frustum,  is  equal  to  the  product  of  half  the  slant-height 
into  the  sum  of  the  perimeters  of  the  ends. 

Ex.  If  the  slant-height  of  a  frustum  of  a  regular  octagonal 
pyramid  be  42  feet,  the  sides  of  one  end  5  feet  each,  and 
the  sides  of  0ie  other  end  3  feet  each  ;  what  is  the  lateral 
surface  ?  Ans.  1344  square  feet. 

52.  If  the  slant-height  be  not  given,  it  may  be  obtained 
from  the  perpendicular  height,  and  the  dimensions  of  the 
two  ends.  Let.  GD  (Fig  17.)  be  the  slant-height  of  the 
frustum  CDGL,  RN  or  GP  the  perpendicular  height,  ND 
and  KG  the  radii  of  the  circles  inscribed  in  the  perimeters 
of  the  two  ends.  Then  PD  is  the  difference  of  the  two 
iljidii :  

And  the  slant-height   GD='^Gir'  -f-PD' 

Ex.  If  the  perpendicular  height  of  a  frustum  of  a  regular 
kexagonal  »yrftaiid  be  24,  the  sides  of  one  end  13  each,  and 
the  sides  of  tke  other  end  8  each ;  what  is  the  whole  surface? 

^WC  -BP'=CP,(Fig.7.)thatis,  ^13^-6:5^  =11.258 

And  '/82--4*=:         6.928 


The  difference  of  the  two  radii  is,  therefore,  4.33 

The  slant-height    =  ^^24' 4-4133^      =24.3875 
The  lateral  surface  is  1536.4 

And  the  whole  surface,  2141.75 

53.  The  height  of  the  whole  pyramid  may  be  calculated 
from  the  dimensions  of  the  frustum.  Let  VN  (Fig,  17.)  be 
the  height  of  the  pyramid,  RN  or  GP  the  height  of  the  frus- 
tum, WD  and  RG  the  radii  of  the  circles  inscribed  in  the 
perimeters  of  the  ends  of  the  frustum. 

Then,  in  the  similar  triangles  GPD  and  VND, 
DP:GP::DN:  VN. 

The  height  f  the  frustum  subtracted  from  VN,  gives  VR 
the  height  of  the  small  pyramid  VLG.  ,  The  solidity  and 
lateral  surfaceof  ihe  frustum  may  then  be  found,  by  subtract- 
ing from  the  whole  pyramid,  the  part  which  is  above  the  cut* 


SOLIDS.  33 

ting  plane.     This  method  may  serve  to  verify  the  calculations 
which  are  made  by  the  rules  in  arts.  50  and  51. 

Ex.  Ifoneendof  fhe  frustum  CDGL  (Fig.  17.)  be:90 
feet  square,  the  other  end  60  feet  square,  and  the  height  RN 
36  feet ;  what  is  the  height  of  the  whole  pyramid  VCD:  and 
what  are  the  solidity  and  lateral  surface  of  the  frustum  ? 

DP=DN-  GR=45-30=15.       And  GP=RN=36. 

Thenl5:36::45:108=VN,theheIghtofthewholepyramid. 
And   108— 36=72=VR,  the  height  of  the  part  VLG. 
The  solidity  of  the  large  pyramid  is  291600    (Art.  48.) 
of  the  small  pyramid        86400 


of  the  frustum  CDGL  205200 


The  lateral  surface  of  the  large  pyramid  is    21060    (Art.  49.) 
of  the  small  pyramid        9360 


of  the  frustum  11700 


PROBLEM    VII. 

To  find  the  solidity  of  a  wedge. 

54.  Add  the  length  of  the  edge  to  twice  the  length  of  the 
base,  and  multiply  the  sum  by  ^  of  the  product  of  the  height 
of  the  wedge  and  the  breadth  of  the  base. 

Let  L=AB  the  length  of  the  base.  (Fig.  20.) 
/=GHthe  length  of  the  edge. 
i  =  BC  the  breadth  of  the  base. 
A=PG  the  height  of  the  wedge. 
Then  L-Z=AB-GH=AM. 

If  the  length  of  the  base  and  the  edge  be  equals  as  BM 
and  GH,  (Fig.  20.)  the  wedge  MBHG  is  half  a  parallelo- 
piped  of  the  same  base  and  height.  And  the  solidity  (Art. 
43.)  is  equal  to  half  the  product  of  the  height,  into  the  length 
and  breadth  of  the  base  ;  that  is  to  i  hhl. 

If  the  length  of  the  base  be  greater  than  that  of  the  edge, 
as  ABGH ;  let  a  section  be  made  by  the  plane  GMN,  par- 


38  MENSURATION  OF 

allel  to  HBC.  This  will  divide  the  whole  wedge  into  two 
pans  MBHG  and  AMG.  The  latter  is  a  pyramid,  whose 
solidity  (Art.  48.)  is  ^bhx{h-^l) 

The  solidity  of  the  parts  together,  is,  therefore, 
ibhl+^,bh  X  (L-Z)=i6A3/-f  iftA2L  -  ibh2l=ibh  x (2L-h/) 
If  the  length  of  the  base  be  less  than  that  of  the  edge,  it 
is  evident  that  the  pyramid  is  to  be  subtracted  from  half  a 
parallelopiped,  which  is  equal  in  height  and  breadth  to  the 
wedge,  and  equal  in  length  to  the  edge. 

The  solidity  of  the  wedge  is,  therefore, 
ibhl-^bhx{l'-h)=lbhSl-}bh2l-\-ibh2h=ibkx(2h-\-l) 

Ex.   1.  If  the  base  of  a  wedge  be  35  by  15,  the  edge  S5, 
and  the  perpendicular  height  12.4;  what  is  the  solidity  ? 

15X12.4 
Ans.  (70+55)  X g— =3875. 

2.  If  the  base  of  a  wedge  be  27  by  8,  the  edge  36,  and  the 
]l«rpendicular  height  42  ;  what  is  the  solidity  ? 

Ans.    5040. 


PROBLEM    VIII. 

To  find  the  solidity  of  a  rectangular  prismoid. 

»5.  To  THE  AREAS  OF  THE  TWO  ENDS,  ADD  FOUR  TIMES  THE  AREA 
OF  A  PARALLEL  SECTION  EQUALLY  DISTANT  FROM  THE  ENDS,  AND  MUL- 
TIPLY THE  SUM  BY  }  OF  THE  HEIGHT. 

Let  L  and  B  (Fig.  21.)  be  the  length  and  breadth  of  one  end, 
I  and  b         the  length  and  breadth  of  the  other  end, 
M  and  m       the  length  and  breadth  of  the  section  in  the 
middle. 
And  h       the  height  of  the  prismoid. 

The  solid  may  be  divided  into  two  wedges,  whose  bases 
are  the  ends  of  the  prismoid,  and  whose  edges  are  L  and  /. 
The  solidity  of  the  whole  by  the  preceding  article,  is 

iBAx(2L-h/)-fi6;iX(2/+L)=iA(2BL-fB/-f2ft/H-JL) 
As  M  is  equally  distant  from  L  and  I, 

2M=L-fZ,2m=B+6,and4Mm=(L-fO(B-f6)=-BL-fBZ-f- 

[bh-\-lb. 


SOLIBS.  39 

Substituting  4M»n  for  its  value  in  the  preceding  expression 
for  the  solidity,  we  have 

i/i(BL-f6/+4M»i) 

That  is,  the  solidity  of  the  prismoid  is  equal  to  }  of  the 
height,  multiplied  into  the  areas  of  the  two  ends,  and  4  times 
the  area  of  the  section  in  the  middle. 

This  rule  may  be  applied  to  prismoids  of  other  forms. 
For,  whatever  be  the  figure  of  the  two  ends,  there  may  be 
drawn  in  each,  such  a  number  of  small  rectangles,  that  the 
sum  of  them  shall  differ  less,  than  by  any  given  quantity, 
from  the  figure  in  which  they  are  contained.  And  the  solids 
between  these  rectangles  will  be  rectangular  prismoids. 

Ex.  1.  If  one  end  of  a  rectangular  prismoid  be  44  feet  by 
23,  the  other  end  36  by  21,  and  the  perpendicular  height  72; 
what  is  the  solidity  ? 

The  area  of  the  larger  ^nd  =44x23=1012 

of  the  smaller  end        =36x21=  756 
of  the  middle  section  =40x22=  880 
And  the  solidity  =(10124-756  +  4x880)  X  12=63456   feet. 

2.  What  is'the  solidity  of  a  stick  of  hewn  timber,  whose  ends 
are  30  inches  by  27,  and  24  by  18,  and  whose  length  is  48 
feet  ?  Ans.  204  feet. 

Other  solids  not  treated  of  in  this  section,  if  they  be  bounded 
by  plane  surfaces,  may  be  measured  by  supposing  them  to  be 
divided  into  prisms,  pyramids,  and  wedges.  And,  indeed, 
every  such  solid  may  be  considered  as  made  up  of  triangular 
pyramids. 


40  MENSURATION   OF 


THE    FIVE    REGULAR    SOLIDS. 

56.  A  SOLID  IS  SAID  TO  BE  REGULAR,  WHEN  ALL  ITS  SOLID  ANGLE? 
ARE  EQUAL,  AND  ALL  ITS  SIDES  ARE  EQUAL  AND  REGULAR  POLYGONS. 

The  following  figures  are  of  this  description  ; 

1.  The   Tetrnedron  ^  f  four  triangles ; 

2.  The  Hexaedron  or  cube  |       u  1  six  squares  ; 

3.  The   Octaedron  )-ZeTlr9  '^  ^'8*"  triangles; 

4.  The  Dodecaedron  I  1  twelve  pentagons; 

5.  The  Icosaedron  J  (^twenty  triangles.* 

Besides  these  five,  there  can  be  no  other  regular  solids. 
The  only  plane  figures  which  can  form  such  solids,  are  tri- 
angles, squares,  and  pentagons.  For  the  plane  angles  which 
contain  any  solid  angle,  are  together  less  than  four  right  an- 
gles or  360°.  (Sup.  Euc.  21.  2.)  And  the  least  number 
which  can  form  a  solid  angle  is  three.  (Sup.  Euc.  Def.  8.  2.) 
If  they  are  angles  of  equilateral  triangles,  each  is  GO''.  The 
sum  of  three  of  them  is  180°,  o(  four  240°,  of^ue  300°,  and 
of  six  360°.  The  latter  number  is  too  great  for  a  solid 
angle. 

The  angles  of  squares  are  90°  each.  The  sum  of  three 
of  these  is  270°,  of  four  360°,  and  of  any  other  greater  num- 
ber still  more. 

The  angles  of  regular  joen^florons  are  108°  each.  The  sum 
o{  three  of  them  is  324°;  of  four,  or  any  other  greater  num- 
ber, more  than  360°.  The  angles  of  all  other  regular  poly- 
gons are  still  greater. 

In  a  regular  solid,  then,  each  solid  angle  must  be  contained 
by  three,  four,  or  five  equilateral  triangles,  by  three  squares, 
or  by  three  regular  pentagons. 

57.  As  the  sides  of  a  regular  solid  are  similar  and  equal, 
and  the  angles  are  elso  alike ;  it  is  evident  that  the  sides  are 
ttU  equally  distant  from  a  central  point  in  the  solid.  If  then, 
planes  be  supposed  to  proceed  from  the  several  edges  to  the 
centre,  they  will  divide  the  solid  into  as  many  equal  pyra- 
mids, as  it  has  sides.  The  base  of  each  pyramid  will  be  one 
of  the  sides  ;  their  common  vertex  will  be  the  central  point ; 
sgnd  their  height  will  be  a  perpendicular  from  the  centre  to 

one  of  the  sides. 

*=  For  the  geometrical  construction  of  these  solids,  see  Legendre's  Geome- 
try ;  Appendix  to  Books  vi  and  vir. 


REGULAR  SOLIDS.  41 


PROBLEM    IX. 

To  find  the  surface  of  a  regular  sq-lid. 

58.  Multiply  the  area  of  one  of  the  sides  bv  tAe  nuhb£<». 
OF  sides.  Or, 

Multiply  the  square  of  oNt  of  THfi  edges,  by  the  surfac* 
OF  a  similar  solid  whose  edges  are   !. 

As  all  the  sides  are  equals  it  is  evident  that  the  area  of  one 
of  them  multiplied  by  the  number  of  sides,  will  give  the  area 
of  the  whole. 

Or,  if  a  table  is  prepared,  containing  the  surfaces  of  the 
several  regular  solids  whose  linear  edges  are  unity  ;  this  may 
be  used  for  other  regular  solids,  upon  the  principle,  that  the 
areas  of  similar  polygons  are  as  the  squares  of  their  homolo- 
gous sides.  (Euc.  20,  6.)  Such  a  table  is  easilv  formed,  by 
multiplying  the  area  of  one  of  the  sides,  as  given*fc  art.  17,  by 
the  number  of  sides.  Thus  the  area  of  an  equilateral  tri- 
angle whose  side  is  1,  is  0.4330127.     Therefore  the  surface 

Of  a  regular  tetraedron  =.4330127  X4=  1.7320508. 
Of  a  regular  octaedron  =.4330127  xS=i=  3.4641 01 6. 
Of  a  regular  icosaedron  =.4330127X20=8.6602540. 

See  the  table  in  the  following  article. 

Ex.  1.  What  is  the  surface  of  a  regular  dodecaedron 
whose  edges  are  each  25  inches.^ 

The  area  of  one  of  the  sides  is  1075.3. 
And  the  surface  of  the  whole  solid   =1075.3X12=12903.6. 

2.  What  is  the  surface  of  a  regular  icosaedron  whose  edges 
are  each  102?  Ans.  90101.3. 


PROBLEM    X. 

To  find  the  solidity  of  a  regular  solid. 
59.  Multiply  the  surface  bv  i  of  the  perpendicular  distance 

FROM  the  centre  TO  ONE  OF    THE  SIDES. 

Or, 
Multiply  the  cube  of  one  of  the  edges,  by  the  solidity  of  a 
similar  solid  whose  edges  are  1 . 

As  the  solid  is  made  up  of  a  number  of  equal  pyramids, 
whose  bases  are  the  sides,  and  whose  height  is  the  perpendic- 

7 


42 


MENSURATION^  OF  REGULAR  SOLIDS. 


ular  distance  of  the  sides  from  the  centre;  (Art.  5?.)  the 
solidity  of  the  whole  must  be  equal  to  the  areas  of  all  the 
sides,  muitiplied  into  i  of  this  perpendicular.  (Art.  48  ) 

If  the  contents  of  the  several  regular  solids  whose  edges 
are  1,  be  inserted  in  a  tabhy  this  may  be  used  to  measure 
other  similar  solids.  For  two  similar  regular  solids  contain 
the  same  number  of  similar  pyramids  ;  and  these  are  to  each 
other  as  the  cubes  of  their  linear  sides  or  edges.  (Sup.  Euc. 
15.  3.  Cor.  3.) 


A  TABLE  OF  REGULAR  SOLIDS  WHOSE  EDGES  ARE  1. 


J^^am^s. 

A*o.  of  sides'. 

Surfaces. 

Solidities. 

Tetraedron 

Hexaedron 

Octaec^rou 

Dodecaedron 

Icosaedron 

4 
6 
8 

12 
20 

1.7320508 
6.0000000 
3.4641016 
20.6457288 
8.6602540 

0.1178513 
1.0000000 
0.4714045 
7.663U89 
2.1816950 

For  the  method  of  calculating  the  last  column  of  this  table, 
see  Hutton's  Mensuration,  Part  IlL  Sec.  2. 

Ex.  What   is  the  solidity  of  a  regular  octaedron  whose 
edges  are  each  32  inches  ?  Ans.  15447  inches. 


SECTION  IT*. 


THE  CYLINDER,  CONE,  AND  SPHERE, 


Abt.  61.     DEr.MT.0.  I.   A^'<^f/,C^Y-r^^'|'' 

a    solid   described    by  the 

revolution  of  a  rectangle  about  one  of  its  sides.  The  ends  or 
bases  are  evidently  equal  and  parallel  circles.  And  the  axis, 
which  is  a  line  passing  through  the  middle  of  the  cylinder,  is 
perpendicular  to  the  bases. 

The  ends  of  an  oblique  cylinder  are  also  equal  and  paral- 
lel circles  ;  but  they  are  not  perpendicular  to  the  axis  The 
height  of  a  cylinder  is  the  perpendicular  distance  from  one 
base  to  the  plane  of  the  other.  In  a  right  cylinder,  it  is  the 
length  of  the  axis. 

II.  A  right  cone  is  a  solid  described  by  the  revoluwon  of  a 
right  angled  triangle  about  one  of  the  sides  which  contaiii  the 
right  angle.  The  base  is  a  circle,  and  is  perpendicular  to  the 
axis,  which  proceeds  from  the  middle  of  the  base  to  the 
vertex. 

The  base  of  an  oblique  cone  is  also  a  circle,  but  is  not  per- 
pendicular to  the  axis.  The  heis:ht  of  a  cone  is  the  perpen- 
dicular distance  from  the  vertex  to  the  plane  of  the  base. 
In  a  right  cone,  it  is  the  length  of  the  axis.  The  slant-height 
of  a  right  cone  is  the  distance  from  the  vertex  to  the  circum- 
ference of  the  base. 

III.  K  frustum  of  a  cone  is  a  portion  cut  off,  by  a  plnne 
parallel  to  the  base.  The  height  of  the  frustum  is  the  per- 
pendicular distance  of  the  two  ends.  The  slant-height  of  a 
frustum  of  a  right  cone,  is  the  distance  between  the  periphe- 
ries of  the  two  ends,  measured  on  the  outside  of  the  solid; 
as  AD.  (Fig.  23.) 

IV.  A  sphere  or  globe  is  a  solid  which  has  a  centre  equally 
distant  from  every  part  of  the  surface.  It  may  be  described 
by  the  revolution  of  a  semicircle  about  a  diameter.  A 
radius  of  the  sphere  is  a  line  drawn  from  the  centre  to  any 

*  Hatton's  Mensuration,  West's  Mathematics,  Legendre's,CIairaut»s,  and 
Camus's  Geometry. 


44  MENSURATION  OF 

part  of  the  surfiice.  A  diameter  is  a  line  passing  through  the 
centre,  and  terminated  at  both  ends  by  the  surface.  The 
circumference  is  the  same  as  the  circumference  of  a  circle 
whose  plane  passes  through  the  centre  of  the  sphere.  Such 
a  circle  is  called  a  great  circle. 

V.  A  segment  of  a  sphere  is  a  part  cut  off  by  any  plane. 
The  height  of  the  segment  is  a  perpendicular  from  the  mid- 
dle of  the  base  to  the  convex  surface,  as  LB.  (Fig.  12.) 

VI.  A  spherical  zone  or  frustum  is  a  part  of  the  sphere  in- 
cluded between  two  parallel  planes.  It  is  called  the  middle 
zone,  if  the  planes  are  equally  distant  from  the  centre.  The 
height  of  a  zone  is  the  distance  of  the  two  planes,  as  LR. 
(Fig.  12.*) 

VII.  A  spherical  sector  is  a  solid  produced  by  a  circular 
sector,  revolving  in  the  same  manner  as  the  semicircle  which 
describes  the  whole  sphere.  Thus  a  spherical  sector  is  de- 
scribed by  the  circular  sector  ACP  (Fig.  15.)  or  GCE  re- 
volving on  the  axis  CP. 

VIII.  A  solid  described  by  the  revolution  of  any  figure 
about  a  fixed  axis,  is  called  a  solid  of  revolution. 


PROBLEM    1. 

To  find  the  coavex  surface  of  a  right  cylinder 

62.    Mui^TIPLY  THE  LENGTH  INTO  THE  CIRCUMFERENCE  OF  THE  BASE. 

If  a  right  cylinder  be  covered  with  a  thin  substance  like 
paper,  which  can  be  spread  out  into  a  plane ;  it  is  evident 
that  the  plane  will  be  a  parallelogram,  whose  length  and 
breadth  will  be  equal  to  the  length  and  circumference  of  the 
cylinder.  The  area  must,  therefore,  be  equal  to  the  length 
multiplied  into  the  circumference.  (Art.  4.) 

Ex.  1  What  is  the  convex  surface  of  a  right  cylinder  which 
is  42  feet  long,  and  15  inches  in  diameter  .'* 

Ans.  42X1.25X3.14159  =  164.938  sq.  feet. 

*  According  to  some  writers,  a  spherical  segment  is  either  a  solid  which  is 
cut  off  from  a  sjjhcro  by  a  single  plaiie,  or  one  which  is  is  included  between 
two  planes :  and  a  zone  the  surface  of  either  of  these.  In  this  sense,  the 
term  zone  is  commonly  used  in  geography. 


THE  CYLINDER.  45 

2  'What  is  the  whole  surface  of  a  right  cylinder,  which   is 
2  feet  in  diameter  and  36  feet  long  ? 

The  convex  surface  is  226. 1 945 

The  area  of  the  two  ends  (Art.  30.)  is       6.2832 


The  whole  surface  is  232.4777 

3.  What  is  the  whole  surface  of  a  right  cylinder  whose 
axis  is  82,  and  circumference  71  ?  Ans.  6624.32. 

63.  It  will  be  observed  that  the  rules  for  the  prism  and 
pyramid  in  the  preceding  sectron,  are  substantially  the  same, 
as  the  rules  for  the  cylinder  and  cone  in  this.  There  may  be 
some  advantage,  however,  in  considering  the  latter  by  them- 
selves. 

In  the  base  of  a  cylinder,  there  may  be  inscribed  a  polygon, 
which  shall  differ  from  it  less  than  by  any  given  space.  (Sup. 
Enc.  6.  1.  Cor.)  If  the  polygon  be  the  base  of  a  prism,  of 
the  same  height  as  the  cylinder,  the  two  solids  may  differ  less 
than  by  any  given  quantity.  In  the  same  manner,  the  base 
of  a  pyramid  may  be  a  polygon  of  so  many  sides,  as  to  differ 
less  than  by  any  given  quantity,  from  the  base  of  a  cone  in 
which  it  is  inscribed.  A  cylinder  is  therefore  considered  by 
many  writers,  as  a  prism  of  an  infinite  number  of  sides  •  and 
a  cone,  as  a  pyramid  of  an  infinite  number  of  sides.  For  the 
meaning  of  the  term  "  infinite,"  when  used  in  the  mathemati- 
cal sense,  see  Alg.  Sec.  xv. 

PROBLEH    II, 

To  find  the  solidity  of  a  cylinder. 

64.  Multiply  the  area  of  the  base  by  the  height. 

The  solidity  of  a  paraUelopiped  is  equal  to  the  product  of 
the  base  into  the  perpendicular  altitude.  (Art.  43.)  And  a 
parallelepiped  and  a  cylinder  which  have  equal  bases  and  alti- 
ludes  are  equal  to  each  other.  (Sup.  Euc.  17.  3.) 

Ex.  1.  What  is  the  solidity  of  a  cylinder,  whose  height  is 
121,  and  diameter  45  2.^ 

Ans.  45.2' X. 7854X121  =  194156.6. 


40  MENSURATION  OF 

2.  What  is  the  soHdity  of  a  cylinder  whose  height  is  424. 
and  circumferenee  213?  Ans.   1530837. 

3.  If  the  side  AC  of  an  oblique  cylinder  (Fig.  22.)  be  27, 
and  the  area  of  the  base  32.61,  and  if  the  side  make  an  angle 
of  62°  44'  with  the  base,  what  is  the  solidity  ? 

R  :  AC : :  Sin  A  :  BC =24  the  perpendicular  height. 
And  the  solidity  is  782.64. 

4.  The  Winchester  bushel  is  ahollow  cylinder,  18^  inches 
in  diameter,  and  8  inches  deep.     What  is  its  capacity .'' 

The  area  of  the  base=(18.5)2X. 7853982=268.8025. 
And  the  capacity  is  2150.42  cubic  inches.  See  the  table 
in  art.  42. 


FUOBLEM    III. 

To  find  the  convex  surface  of  a  right  cone. 

65.  Multiply  half  the  slant-height  into  the  circumference 
of  the  base. 

If  the  convex  surface  of  a  right  cone  be  spread  out  into  a 
plane,  it  will  evidently  form  a  sector  of  a  circle  whose  radius 
is  equal  to  the  slant-height  of  the  cone.  But  the  area  of  the 
sector  is  equal  to  the  product  of  half  the  radius  into  the 
length  of  the  arc.  (Art.  34.)  Or  if  the  cone  be  considered 
as  a  pyramid  of  an  infinite  number  of  sides,  its  lateral  surface 
is  equal  to  the  product  of  half  the  slant-height  into  the  perim- 
eter of  the  base.  (Art.  49.) 

Ex.  1.  If  the  slant-height  of  a  right  corie  be  82  feet,  and 
the  diameter  of  the  base  24,  what  is  the  convex  surface  ? 
Ans.  41  X2ix3. 14159  =  3091. 3  square  feet. 

2.  If  the  axis  of  a  right  cone  be  48,  and  the  diameter  of 
the  base  72,  what  is  the  whole  surface  ^ 

The  slant-height  ='^36'-f48''=60.  (Euc.  47.  1.) 
The  convex  surface  is  678G 

The  area  of  the  base  4071.6 

And  the  whole  surface  10857.6 


THE  CONE.  47 

3.  If  the  axis  of  a  right  cone  be   16,  and  the  circumfer- 
ence of  the  base  75.4 ;  what  is  the  whole  surface  ? 

Ans.   1206.4. 


PROBLEM    IV. 

To  find  the  solidity  of  a  cone. 

60.  Multiply  the  area  of  the  base  into  ^  of  the  height. 

The  solidity  of  a  cylinder  is  equal  to  the  product  of  the 
base  into  the  perpendicular  height.  (Art.  64.)  And  if  a  cone 
and  a  cylinder  have  the  same  base  and  altitude,  the  cone  is 
i  of  the  cylinder.  (Sup.  Euc.  18.  3.)  Or  if  a  cone  be  con- 
sidered as  a  pyramid  of  an  infinite  number  of  sides,  the  solid* 
ity  is  equal  to  the  product  of  the  base  into  \  of  the  height,  by 
art.  48. 

Ex.  1.  What  is  the  solidity  of  a  right  cone  whose  height  is 
663,  and  the  diameter  of  whose  base  is  101  ^ 

Ans.    foT^  X  .7854  X  221  =  1770622. 

2.  If  the  axis  of  an  oblique  cone  be  738,  and  make  an 
angle  of  30°  with  the  plane  of  the  base  ;  and  if  the  circumi- 
ference  of  the  base  be  355,  what  is  the  solidity  ^ 

Ans.  1233536. 


PROBLEM    V. 

To  find  the  convex  surface  of  a  frustum  of  a  right  cone, 

67.  Multiply  half  the  slant-height  by  the  sum  of  the  pe*- 
ripheries  of  the  two  ends. 

This  is  the  rule  for  a  frustum  of  a  pyramid;  (Art.  51.)  and 
is  equally  applicable  to  a  frustum  of  ^cone,  if  a  cone  be  con- 
sidered as  a  pyramid  of  an  infinite  number  of  sides.  (Art.  63.) 

Or  thus, 

Let  the  sector  ABV  (Fig.  23.)  represent  the  convex  sur- 
face of  a  right  cone,  (Art.  65.)  and  DCV  the  surface  of  a 
portion  of  the  cone,  cut  off  by  a  plane  parallel  to  the  base. 
Then  will  ABCD  be  the  surface  of  the  frustum. 


48  MENSURATION  OF 

Let  AB=fl,        DC=6,  VD=rf,        AD=A, 

Then  the  area  ABV =iax{h-^d)==^ah+^ad.    (Art.  34.) 
And  the  area  DCV=i*rf. 

Subtracting  the  one  from  the  other, 

The  area  ABDC=iaA-f  iarf-|6(^. 

But  d :  d+h::b  :a.  (Sup.Euc.8.1  )  Therefore  iad^ibd=^ibh. 

The  surface  of  the  frustuno,  then,  is  equal  to 
iflA-f  16^.  or  lAx(a-l-ft) 

Cor.  The  surface  of  the  frustum  is  equal  to  the  product  of 
the  slant-height  into  the  circumference  of  a  circle  which  is 
equally  distant  from  the  two  ends.  Thus  the  surface  ABCD 
(Fig.  23.)  is  equal  to  the  product  of  AD  into  MN.  For  MN 
is  equal  to  half  the  sum  of  AB  and  DC. 

Ex.  1.  What  is  the  convex  surface  of  a  frustum  of  a  right 
cone,  if  the  diameters  of  the  two  ends  be  44  and  33,  and  the 
slant-height  84  f  Ans.  1 0 1 59.8. 

2.  If  the  perpendicular  height  of  a  frustum  of  a  right  cone 
be  24,  and  the  diameters  of  the  two  ends  80  and  44,  what  is 
the  whole  surface  ? 

Half  the  difference  of  the  diameters  is  18. 

And  ^18' -f- 24' =30,  the  slant-height,  (Art.  52.) 
The  convex  surface  of  the  frustum  is  5843 

The  sum  of  the  areas  of  the  two  ends  is  6547 

And  the  whole  surface  is  12390 


PROBLEM    VI. 

To  find  the  solidity  of  a  frustum  of  a  cone» 
68.   Add  together  the  areas  of  the  two  ends,  and  the  square 

root  of  the  product  of  these  areas  ;  AND  MULTIPLY  THE  SUM  BY 
L  OF  THE  PERPENDICULAR  HEIGHT. 

This  rule,  which  was  given  for  the  frustum  of  a  pyramid^ 
(Art,  50.)  K  equally  applicable  to  the  frustum  of  a  cone  ;  be- 
cause a  cone  and  a  pyramid  which  have  equal  bases  and  alti- 
tudes are  equal  to  each  other. 


THE  SPHERE.  49. 

Ex.  1.  What  is  the  solidity  of  a  mast  which  is  72  feet 
long,  2  feet  in  diameter  at  oae  end,  and  18  inches  at  the 
other  ?  "  Ans.  174.36  cubic  feet. 

2.  What  is  the  capacity  of  a  conical  cistern  which  is  9  feet 
deep,  4  feet  in  diameter  at  the  bottom,  and  3  feet  at  the  top? 

Ans.  87.18  cubic  feet  =652.15  wine  gallons. 

3.  How  many  gallons  of  ale  can  be  put  into  a  vat  in  the 
form  of  a  conic  frustum,  if  the  larger  diameter  be  7  feet,  the 
smaller  diameter  6  feet,  and  the  depth  8  feet  ? 


PROBLEM    Vlt.  .    •     '"^ 

To  find  the  surtace  of  a  sphere. 

69.  Multiply  the  diameter  by  the  circumference. 

Let  a  hemisphere  be  described  by  the  quadrant  CPD, 
(Fig.  25.)  revolving  on  the  line  CD.  Let  AB  be  a  side  of  a 
regular  polygon  inscribed  in  the  circle  of  which  DBP  is  an 
arc.  Draw  AO  and  BN  perpendicular  to  CD,  and  BH  per- 
pendicular to  AO.  Extend  AB  till  it  meet  CD  continued. 
The  triangle  AOV,  revolving  on  OV  as  an  axis,  will  describe 
a  right  cone.  (Defin.  2.)  AB  will  be  the  slant-height  of  a 
frustum  of  this  cone  extending  from  AO  to  BN.  From  G 
the  middle  of  AB,  draw  GM  parallel  to  AO.  The  surface 
of  the  frustum  described  by  AB,  (Art.  67.  Cor.)  is  equal  to 

ABxctVcGM.* 

From  the  centre  C  draw  CG,  which  will  be  perpendicular 
to  AB,  (Euc.  3.  3.)  and  the  radius  of  a  circle  inscribed  in  the 
polygon.  The  triangles  ABH  and  CGM  are  similar,  be- 
cause the  sides  are  perpendicular,  each  to  each.     Therefore, 

HB  or  ON  :  AB:  :GM  :  GC:  .arc  GM  :  circ  GC. 

So  that  ON  X  circ  G€=ABxciVc  GM,  that  is,  the  surface 
of  the  frustum  is  equal  to  the  product  of  ON  the  perpendicu- 
lar height,  into  circ  GC,  the  perpendicular  distance  from  thb 
centre  of  the  polygon  to  one  of  the  sides. 

*  By  circ  GM  is  meant  the  circumference  of  a  eirel«  the  rjdiu?  of  yrhii^  h 
GM. 

8 


5b  MENSURATION  OF 

In  the  same  manner  it  may  be  proved,  that  the  surfaces 
produced  by  the  revolution  of  the  lines  BD  and  AP  about 
the  axis  DC,  are  equal  to 

NDX  arc  GC,  and  CO  Xcirc  GC. 

The  surface  of  the  whole  solid,  therefore,  (Euc.l.2.)is  equal  to 
CD  xcirc  GC. 

The  demonstration  is  applicable  to  a  solid  produced  by 
the  revolution  of  a  polygon  of  any  number  of  sides.  But  a 
polygon  may  be  supposed  which  shall  differ  less  than  by  any 
given  quantity  from  the  circle  in  which  it  is  inscribed  ;  (Sup. 
Euc.  4.  I.)  and  in  which  the  perpendicular  GC  shall  differ 
less  than  by  any  j^iven  quantity  from  the  radius  of  the  circle. 
Therefore  the  surface  of  a  hemisphere  is  equal  to  the  product 
of  its  radius  into  the  circumference  of  its  base;  and  the  sur- 
face of  a  :tpkere  i$  equal  to  the  product  of  its  diameter  into  its 
circumference. 

Cor,  1.  From  this  demonstration  it  follows,  that  the  sur- 
face of  any  ^g>nent  or  zone  of  a  sphere  is  equal  to  the  pro- 
duct of  the  height  of  the  segment  or  zone  into  the  circum- 
ference of  the  sphere.  The  surface  of  the  zone  produced 
by  the  revolution  of  the  arc  AB  about  ON,  is  equal  ta 
ON  xc/rc  CP.  And  the  surface  of  the  segment  produced 
by  the  rerolution  of  BD  about  DN  is  equal  to  DN  XaVc  CP, 

Cor.  2.  The  surface  of  a  sphere  is  equal  to  four  times  the 
area  of  a  circle  of  the  same  diameter;  and  therefore,  the 
convex  surface  of  a  hemisphere  is  equal  to  twice  the  area  of 
its  base.  For  the  area  of  a  circle  is  equal  to  the  product  of 
half  the  diameter  into  half  the  circumference  ;  (Art.  30.)  that 
is,  to  J  the  product  of  the  diameter  and  circumference. 

Cor.  3.  The  surface  of  a  sphere,  or  the  convex  surface  of 
any  spherical  segment  or  zone,  is  equal  to  that  of  the  cir- 
cumscribing cylinder.  A  hemisphere  described  by  the  rev- 
olution of  the  arc  DBP,  is  circumscribed  by  a  cylinder  pro- 
duced by  the  revolution  of  the  parallelogram  D^CP,  The 
convex  surface  of  the  cylinder  is  equaf  to  its  height  multi- 
plied by  its  circu.jiference.  (Art.  62.)  And  this  is  also,  the 
surface  of  the  hemisphere. 


THE  SPHERE.  51 

So  the  surface  produced  by  the  revolution  of  AB  is  equal 
to  that  produced  by  the  revolu'ion  of  ab.  And  the  surface 
produced  by  BD  is  equal  to  that  produced  by  hd, 

Ex.  1.  Considering  the  earth  as  a  sphere  7930  miles  in 
diameter,  how  manv  square  miles  are  there  on  its  surface  ? 

Ans.  197,558,500. 

2.  If  the  circumference  of  the  sun  be  2.800,000  miles, 
what  is  his  surface  i*        Ans.  2,495,547,600,000  sq.  miles. 

3.  How  many  square  feet  of  lead  will  it  require,  to  cover  a 
hemispherical  dome  whose  base  is  13  feet  across  ^ 

Ans.  2651. 


PROBLEM    VIII. 

To  find  the  solidity  of  a  sphsrb. 

70.     I4  Multiply  the  cube  op  the  diameter  by  .O^ot>« 

Or, 

2.  Multiply  the  squ.ire  of  the  diameter  by  J-  of  the  ctR- 

CCMFERENCE.  Of, 

3.  Multiply  the  surface  by  |  of  the  diameter. 

1.  A  sphere  is  two  thirds  of  its  circumscribing  cylinder. 
(Sup.  Euc.  21.  3.)  The  height  and  diameter  of  the  cylinder 
are  each  equal  to  the  diameter  of  the  sphere.  The  solidity 
of  the  cylinder  is  equal  to  its  height  multiplied  into  the  area 
of  its  base,  (Art.  64.)  that  is  putting  D  for  the  diameter, 

DXD2X.7854     or     D3X.7854. 

And  the  solidity  of  the  sphere,  being  |  of  this,  is 
D3X.5236. 

2.  The  base  of  the  circumscribing  cylinder  is  equal  to  hal 
the  circumference  multiplied  into  half  the  diameter,  (Art.  30.) 
that  is,  if  C  be  put  for  the  circumference, 

iC  xD,  and  the  solidity  is  iC  xD^. 

Therefore  the  solidity  of  the  sphere  is 

fof  iCxD2=D2xia 


•b^  ftlENSURATION  OF 

3.  In  the  last  expression,  which  is  the  same  as  C  X  D  X  ^Vr 
we  may  substitute  S,  the  surface,  for  C  xD.  (Art.  69.)  We 
then  have  the  solidity  of  the  sphere  equal  to 

SxiD. 

Or,  the  sphere  may  be  supposed  to  be  filled  with  small 
pyramids,  standing  on  the  surface  of  the  sphere,  and  having 
their  common  vertex  in  the  centre.  The  number  of  these 
may  be  such,  that  the  difference  between  their  sum  and  the 
sphere  shall  be  less  than  any  given  quantity.  The  sotidity  of 
each  pyramid  is  equal  to  the  product  of  its  base  into  ^  of  its 
height.  (Art.  48.)  The  solidity  of  the  whale,  therefore,  h 
equal  to  the  product  of  the  surface  of  the  sphere  into  ^  of  its 
radius,  or  i  of  its  diameter. 

71.  The  numbers  3.14159,  .7854,  .5236,  should  be 
made  perfectly  familiar.  The  first  expresses  the  ratio  of  the 
ctrcvm/ercnce.of  a  circle  to  thcf/iame^er;  (Art.  23.)  the  sec- 
ond, the  ratio  of  the  area  of  a  circle  to  the  square  of  the  diam- 
eter (Art.  30.) ;  and  the  third,  the  ratio  of  the  solidity  of  a 
sphere  to  the  cube  of  the  diameter.  The  second  is  |  of  the 
first,  and  the  third  is  |  of  the  first. 

As  these  numbers  are  frequently  occurring  in  mathematical 
investigations,  it  is  common  to  represent  the  first  of  them  by 
the  Greek  letter  11.     According  to  this  notation, 

*=3.14159,     i*=.7854.     i*=.5236. 

If  D=the  diameter,  and  R=the  radius  of  any  circle  or  sphere; 

Then     D=2R     D2=4R«     D='=8R3. 

And-rrD)        .  i       l-ffDa  >=  the  area  of  I  irD')        , 

Or2^R^  =thepen;,A.  J^j^,  ^   ^^^  ^^^^^    orl.R3l  =^»^^ 

solidity  of  the  sphere, 

Ex.  1-  What  is  the  solidity  of  the  earth,  if  it  be  a  sphere 
7930  miles  in  diameter  ? 

Ans.  261,107,000,000  cubic  miles. 

2.  How  many  wine  gallons  will  fill  a  hollow  sphere  4  feet 
in  diameter  ? 

Ans.  The  capacity  is  33.5104  feet=250|  gallons. 

3.  If  the  diameter  of  the  moon  be  2180  miles,  what  is  its 
4iOlidity  ?  Ans.  5,424,600,000  miles* 


THE   SPHERE.  53 

72.  If  the  solidity  of  a  sphere  he  given,  the  diameter  may 
be  found  b}'  reversing  the  first  rule  in  the  preceding  article; 
that  is,  dividing  by  .5236  and  extracting  the  cube  root  of  the 
quotient. 

Ex  1.  Wh^t  is  the  diameter  of  a  sphere  whose  solidity- 
is  65.45  cubic  feet?  Ans.     5  feet. 

2.  What  must  be  the  diameter  of  a  globe  to  contain 
16755  pounds  of  water?  Ans.     S  feet. 


PROBLEM    IX. 

To  find  the  convex   surface  q/*  a   segment  or  zone    of  a 
^  sphere* 

73.  Multiply  the  height  of  the  segment  or  zonk  into  the  cir- 
cumference OF  THE  sphere. 

For  the  demonstration  of  this  rule,  see  art.  69. 

Ex.  1.  If  the  earth  be  considered  a  perfect  sphere  7930 
miles  in  diameter,  and  if  the  polar  circle  be  23*'  28'  from  the 
pole,  how  many  square  miles  are  there  in  one  of  the  frigid 
zones  ? 

If  PQOE  (Fig.  15.)  be  a  meridian  on  the  earth,  ADB  one 
of  the  polar  circles,  and  P  the  pole;  then  the  frigid  zone  is 
a  spherical  segment  described  by  the  revolution  of  the  arc 
APB  about  PD.  The  angle  ACD  subtended  by  the  arc 
AP  is  23°  28'.     And  in  the  right  angled  triangle  ACD, 

R  :  AC: : Cos  ACD  :  CD=3637. 

Then  CP~CD=3965-3637=328=PD  the  height  of 
the  segment. 

And  328X7930X3.14159  =  8171400  the  surface. 

2.  If  the  diameter  of  the  earth  be  7930  miles,  what  is  the 
surface  of  the  torrid  zone,  extending  23°  28'  on  each  side  of 
the  equator  ? 

If  EQ  (Fig.  15.)  be  the  equator,  and  GH  one  of  the  tro- 
pics, then  th*'  angle  ECG  is  23°  28'.  And  in  the  right  angled 
triangle  GCM,- 


54  MENSURATION  OF 

R  :  CG:  :Sin  ECG  :  GM=CN  =  1678.9  the  height  of  half 
the  zone. 

The  surface  of  the  whole  zone  is  78669700. 

3.     What  is  the  surface  of  each  of  the  temperate   zones  ? 
The  height  DN=CP-CN-PD=^0o8.1 
And  the  surface  of  the  zone  is  51273000. 

The  surface  of  the  two  temperate  zones  is  102,540,000 
of  the  two  frigid  zones  16,342,000 

of  the  torrid  zone  78,669,700 


of  the  whole  globe  1 97, 558,500 

PROBLEM  X. 

To  find  the  solidity  of  a  spherical  sector, 
74.     Multiply  the  spherical  surface  by  ^  of  the  radius  of 

THE  sphere. 

The  spherical  sector,  (Fig.  24.)  produced  by  the  revolu- 
tion of  ACBD  about  CD,  may  be  supposed  to  be  filled  with 
small  pyramids^  standing  on  the  spherical  surface  ADB,  and 
terminating  in  the  point  C  Their  number  may  be  so  great, 
that  the  heightofeach shall  differ  less  than  by  any  given  length 
from  the  radius  CD,  and  the  sum  of  their  bases  shall  differ 
less  than  by  any  given  quantity  from  the  surface  ABD.  The 
tsolidity  of  each  is  equal  to  the  product  of  its  base  into  i  of 
the  radius  CD.  (Art.  48.)  Therefore,  the  solidity  of  all  of 
them,  that  is,  of  the  sector  ADBC,  is  equal  to  the  product  of 
the  spherical  surface  into  j  of  the  radius. 

Ex.  Supposing  the  earth  to  be  a  sphere  7930  miles  in  di- 
ameter, and  the  polar  circle  ADB  (Fig.  15.)  to  be  23°  28' 
from  the  pole ;  what  is  the  solidity  of  the  spherical  sector 
ACBP? 

Ans.     10,799,867,000  miles 


THE  SPHERE.  55 

PROBLEM    XI, 

To  find  the  solidity  of  a  spherical  segment. 

75.  Multiply  half  the  height  of  the  segment  into  the  area 
OF  the  base,  and  the  cube  of  the  height  into  .5236  ;  and  add 
the  two  products. 

As  the  circular  sector  AOBC  (Fig.  9.)  consists  of  two 
parts,  the  segment  AOBP  and  the  triangle  ABC;  (Art.  35.) 
so  the  spherical  sector  produced  by  the  revolution  of  AOC 
about  OC  consists  of  two  parts,  the  segment  produced  by  the 
revolution  of  AOP,  and  the  cone  produced  by  the  revolution 
of  ACP.  If  then  the  cone  be  subtracted  from  the  sector, 
the  remainder  will  be  the  segment. 

Let  CO=R  the  radius  of  the  sphere, 

PB  =  r  the  radius  of  the  base  of  the  segment, 
PO=A  the  height  of  the  segment, 
Then  PC=R  — A  the  axis  of  the  cone. 
The  sector=2'rrR  xhx^R  (Arts.  71,  73,  74.)  =  f -^r/tRs 
The  cone='«'r3  xi(R-A)  (Arts.  71,66.)  =  icrr5^R~^<;r/i;.2. 

Subtracting  the  one  from  the  other, 
The  segment  =t<jrAR» -i-jrrsR^^-^^^s^ 

r,utDOxPO=BO'(Trig.97.*)=PO^+PB^  (Euc.  47.  i.) 

Thatis,  2RA=A3-hr^     So  that,  R= ^^±11 

V    2A     7  4h' 

Substituting  then,  for  R  and  R%  their  values,  and  multi 
plying  the  factors, 

The  segment=iorA'-f-i'^Ar2-f  i'^-^^Ar^-i^+i  hr' 

which,  by  uniting  the  terms,  becomes 
i'^hr^^iifh^, 

*  Euclid  31, 3,  and  8,  6.  Cor. 


5G  MENSURATION  OF 

The  first  term  here  is  i^X^r^,  half  the  height  of  the  seg- 
ment multiplied  into  the  area  of  the  base  5  (Art.  71.)  and  the 
other  h^  X^'f,  the  cube   of  the  height  multiplied  into  .S'iSG. 

If  the  segment  be  .^rcafcr  than  a  hemisphere,  as  ABD  ; 
(Fig.  9.)  the  cone  ABC  must  be  added  to  the  sector  ACBD, 

^^  L?t  PD=A  the  height  of  the  segment, 

Then  PC=A  -  R  the  axis  of  the  cone. 

The  sector  ACBD=|irAR2 

The  cone=^>''  Xi(A-R)=i^Ar«  -i«-r«R 

Adding  them  together,  we  have  as  before, 

The  segment= |^AR3  -  J^r^  R-f  \^hr^ , 

Cor.  The  solidity  of  a  spherical  segment  is  equal  to  half 
a  cylinder  of  the  same  base  and  height -f-a  sphere  whose  di- 
ameter is  the  height  of  the  segment.  For  a  cyHnder  is 
equal  to  its  height  multiplied  into  the  area  of  its  base  ;  and 
a  sphere  is  equal  to  the  cube  of  its  diameter  multiplied  by 
.6236. 

Thus  ifOy  (Fig.  15.)  be  half  Ox,  the  spherical  segment 
produced  by  the  revolution  of  Oxt  is  equal  to  the  cylinder 
produced  by  tvyx  -f  the  sphere  produced  by  Oyxz;  sup- 
posing each  to  revolve  on  the  line  Ox. 

Ex.  1 .  If  the  height  of  a  spherical  segment  be  8  feet,  and 
the  diameter  of  its  base  25  h^i ;  what  is  the  solidity  ? 

Ans.     25'x.7854X4  +  8='X.5236=225l.58feet. 

2.  If  the  earth  be  a  sphere  7930  miles  in  diameter,  and 
the  polar  circle  23°  28'  from  the  pole,  what  is  the  solidity  of 
one  of  the  frigid  zones  ? 

Ans.     I,303,000,.0p0  miles. 


THE  SPHERE.  57 

PROBLEM     XII. 

To  find  the  solidity  of  a  spherical  zose  or frusiunu 

76.  From  the  solidity  of  the  whole  sphere,  subtract  the  two 
segments  on  the  sides  of  the  zone. 

Or, 

Jdd  together  the  squares  of  the  radii  of  the  two  ends^  and  1 
the  square  of  their  distance ;  and  multiply  the  sum  by  three, 
times  this  distance^  and  the  product  by  .5236. 

If  from  the  whole  sphere,  (Fig.  15.)  there  be  taken  the 
two  segments  ABP  and  GHO,  there  will  remain  the  zone  or 
frustum  ABGH. 

Or,  the  zone  ABGH  is  equal  to  the  difference  between  the 
segments  GHP  and  ABP. 

LetNP=H>  ,.     ,   .  ,       .  , 

r)p__  7    >  the  heights  ot  the  two  segments. 

AQ~      >  the  radii  of  their  bases. 

DN=c?=H  — A  the    distcktice  of  the  two  bases, 
or  the  height  of  the  zone. 

Then  the  larger  segment=iTHR2-|-|'?rH='   \  , k^    -.  \ 
And  the  smaller  segment=^'TAr2 +i<jrA3        ^  V^^-  75.) 

Therefore  the  zone  ABGH  =  H3HR2-fH^—3Ar2-A3) 

By  the   properties  of  the   circle,  (Euc.  35,  3.) 
ONxH=R^     Therefore  (ON-hH)  xH=R=+H^ 
R2-I-H2 
Or  0P= — jj 

In  the  same  manner,     0P=— jj^ 


Therefore  3H  X(r«-|-A^)=3/iX(R-^4-H»). 
Or  3Hr2-f3H/i2-3AR2^3H^/i=0.     (Alg.  178.) 


58  MENSURATION  OF 

To  reduce  the  expression  for  the  solidity  of  the  zoue  (• 
the  required  form,  without  altering  its  value,  let  these  terms 
be  added  to  it:  and  it  will  become 

|i^(3HR'+3Hr2-3AR»-3Ar3-f-H3-3H2A-f3HA«-/t») 

Which  is  equal  to 

^-•X3(H-/i)X(R^+r*-|-i(H-A)2) 

Or,  as  jflf  equals  .5236  (Art.  71.)  and  H  —  h  equals  d^ 

The  zone  =  .5236  X Sc^X (R^  +r^  +1^^) 

Ex.  1.  If  the  diameter  of  one  end  of  a  spherical  zone  is 
24  feet,  the  diameter  of  the  other  end  20  feet,  and  the  dis- 
tance of  the  two  ends,  or  the  height  of  the  zone  4  feet ;  what 
is  the  solidity  ?  Ans.     1566.6  feet. 

2.  If  the  earth  be  a  sphere  7930  miles  in  diameter,  and 
the  obliquity  of  the  ecliptic  23°  28';  what  is  the  solidity  of 
one  of  the  temperate  zones  ? 

Ans.     55,390,500,000  miles. 

3.  What  is  the  solidity  of  the  torrid  zone  ? 

Ans.     147,726,000,000  miles. 

yhe  solidity  of  the  two  temperate  zones  is  1 10,781,000,000 
of  the  two  frigid  zones  2,606,000,000 

of  the  torrid  zone  1 47,720,000,000 


of  the  whole  globe  261,107,000,00f^ 


•OLIDS.  *5d 


PROMISCUOUS  EXAMPLES  OF  SOLIDS. 

Ex.  1.  How  much  water  can  be  put  into  a  cubical  ves- 
sel three  feet  deep,  which  has  been  previously  filled  witk 
cannon  balls  of  the  same  size,  2,  4,  6,  or  9  inches  in  diame- 
ter, regularly  arranged  in  tiers,  one  directly  above  another^ 

Ans.     96}  wine  gallons. 

2.  If  a  cone  or  pyramid,  whose  height  is  three  feet,  be 
divided  into  three  equal  portions,  by  sections  parallel  to  the 
base  ;  what  will  be  the  heights  of  the  several  parts  ? 

Ans.     24.961,     6.488,  and  4.551  inches. 

3.  What  is  the  solidity  of  the  greatest  square  prism  which 
can  be  cut  from  a  cylindrical  stick  of  timber,  2  feet  6  inches 
in  diameter  and  56  feet  long  ?* 

Ans.     175  cubic  feet. 

4.  How  many  such  globes  as  the  earth  are  equal  in  bulk 
to  the  sun  ;  if  the  former  is  79S0  miles  in  diameter,  and  the 
latter  890,000  ?  Ans.     1,413,678. 

5.  How  many  cubic  feet  of  wall  are  there  in  a  conical 
tower  66  feet  high,  if  the  diameter  of  the  base  be  20  feet  from 
outside  to  outside,  and  the  diameter  of  the  top  8  feet  ;  the 
thickness  of  the  wall  being  4  feet  at  the  bottom,  and  decreas- 
ing regularly,  so  as  to  be  only  2  feet  at  the  top  ? 

Ans.     7188. 

*  The  common  rule  for  measuring  round  timber  is  to  multiply  the  square  of 
the  quarter-girt  by  the  length.  The  quarter  girt  is  one  fourth  of  the  circum. 
ference.  This  method  does  not  give  the  whole  solidity.  It  makes  an  allow- 
ance of  akKjut  one-fifth,  for  wa?te  in  hewing,  bark,  &c.  The  solidity  of  a  cyl- 
inder is  equal  to  the  product  of  the  length  into  the  area  of  the  base. 

If  C=the  circumference,  andtr=3.14159,  then  (Art.  31.) 

C2         /       C        \3_      /       O       \3 

Theareaofthebasc=^;:^^;^^  j-    ^3;^43-j 

If  then  the  circumference  were  divided  by  3.545,  instead  of  4,  and  the  quo- 
tient squared,  the  area  of  the  base  would  be  correctly  found.    See  note  G. 


g«  MENSURATION  OF  SOLIDS. 

0.  If  a  metallic  globe  is  filled  with  wine,  which  cost  as 
much  at  5  dollars  a  gallon,  as  the  globe  itself  at  20  cents  for 
every  square  inch  of  its  surface  ;  what  is  the  diameter  of  the 
globe  ?  Ans.    55.44  inches. 

7.  If  the  circumference  of  the  earth  be  25,000  miles, 
what  must  be  the  diameter  of  a  metallic  globe,  which,  when 
drawn  into  a  wire  2V  ^^  ^^  ^"^^^  '"  diameter,  would  reach 
round  the  earth  ?  Ans.     15  feet  and  1  inch. 

8.  If  a  conical  cistern  be  3  feet  deep,7i  feet  in  diameter 
at  the  bottom,  and  5  feet  at  the  top  ;  what  will  be  the  depth 
of  a  fluid  occupying  half  its  capacity  ? 

Ans.     14.535  inches. 

9.  If  a  globe  20  inches  in  diameter  be  perforated  by  a 
cylinder  16  inches  in  diameter,  the  axis  of  the  latter  passing 
through  the  centre  of  the  former ;  what  part  of  the  solidity, 
and  the  surface  of  the  globe  will  be  cut  away  by  the  cylin- 
der ? 

Ans.  3284  inches  of  the  solidity,  and  502,655  of  the  surface. 

10.  What  is  the  solidity  of  the  greatest  cube  which  can 
be  cut  from  a  sphere  three  feet  in  diameter  ? 

Ans.     5t  feet 


SECTION  V. 


ISOPERIMETRY.* 


.  --  TT  is  often  necessary  to  connpare  a  number  of 
different  figures  or  solids,  for  the  purpose  of  as- 
certaining which  has  the  greatest  area,  within  a  given  perinn- 
eter,  or  the  greatest  capacity  under  a  given  surface.  We  may 
have  occasion  to  determine,  for  instance,  what  must  be  the 
form  of  a  fort,  to  contain  a  given  number  of  troops,  with  the 
least  extent  of  wall ;  or  what  the  shape  of  a  metallic  pipe  to 
convey  a  given  portion  of  water,  or  of  a  cistern,  to  hold  a 
given  quantity  of  liquor,  with  the  least  expense  of  materials. 

78.  Figures  which  have  equal  perimeters  are  called  Iso- 
perimeters.  When 'a  quantity  is  greater  than  any  other  of  the 
same  class,  it  is  called  a  maximum.  A  multitude  of  straight 
lines,  of  different  lengths,  may  be  drawn  within  a  circle^  But 
among  them  all,  the  diameter  is  a  maximum.  Of  all  sines  of 
angles,  which  can  be  drawn  in  a  circle,  the  sine  of  90"^  is  a 
maximum. 

When  a  quantity  is  less  than  any  other  of  the  same  class, 
it  is  called  a  minimum.  Thus,  of  all  straight  lines  drawn 
from  a  given  point  to  a  given  straight  line,  that  which  is  per- 
jjendicular  to  the  given  line  is  a  minimum.  Of  all  straight 
lines  drawn  from  a  given  point  in  a  circle,  to  the  circumfer- 
ence, the  maximum  and  minimum  are  the  two  parts  of  the 
diameter  which  pass  through  that  point.     (Euc,  7,  3.) 

In  isoperimetry,  the  object  is  to  determine,  on  the  one 
hand,  in  what  cases  the  area  is  a  maximum,  within  a  given 
perimeter;  or  the  capacity  a  maximum,  within  a  given  sur- 
face :  and  on  the  other  hand,  in  what  cases  ihe  perimeter  is 
a  minimum  for  a  given  area,  or  the  surface  a  minimum,  for  a 
given  capacity. 

♦Emerson's,  Simpson's,  and  Legendre's  Geometry,  Lhuillipr.  Fontenelle, 
■  Vg1.'75. 


C5  MENSURATION. 


PROPOSITION  I. 

79.  An  Isosceles  Triangle  has  a  greater  area  than  any- 
scalene  iriangUj  of  equal  base  and  perimeter. 

If  ABC  (Fig.  26.)  be  an  isosceles  triangle  whose  equal 
sides  are  AC  and  BC  ;  and  if  ABC  be  a  scalene  triangle  on 
the  same  base  AB,  and  having  AC'  +  BC  =  AC  +  BC ;  then 
the  area  of  ABC  is  greater  than  that  of  ABC. 

Let  perpendiculars  be  raised  from  each  end  of  the  base, 
extend  AC  to  D,  make  CD'  equal  to  AC,  join  BD,  and 
draw  CH  and  CH'  parallel  to  AB. 

As  the  angle  CAB=ABC,  (Euc.  5,  1.)  and  ABDis  a  right 
angle,  ABC+CBD=CAB-f  CDB=ABC+CDB.  There- 
fore CBD=CDB,  so  that  CD=CB  ;  and  by  construction, 
C'D'=^AC\  The  perpendiculars  of  the  equal  right  angled 
triangles  CHD  and  CHB  are  equal;  therefore  BH  =  iBD. 
In  the  same  manner,  AH'  =  iAD'.  The  line  AD=AC  +  BC 
=AC-i-BC=D'C+BC.  But  D'C+BC>BD'.  (Euc. 
20,1.)  Therefore,  AD>BD';  BD>AD',  (Euc.  47,  1.) 
and  lBD>iAD'.  But  iBD,  or  BH,  is  the  height  of  the 
isosceles  triangle;  (Art.  1.)  and  |AD'  or  AH',  the  height  of 
the  scalene  triangle  ;  and  the  areas  of  two  triangles  which 
have  the  same  base  are  as  their  heights.  (Art.  8.)  There- 
fore the  area  of  ABC  is  greater  than  that  of  ABC.  Among 
all  triangles,  then,  of  a  given  perimeter,  and  upon  a  given 
base,  the  isosceles  triangle  is  a  maximum. 

Cor.  The  isosceles  triangle  has  a  less  perimeter  than  any 
scalene  triangle  of  the  same  base  and  area.  The  triangle 
ABC'  being  less  than  ABC,  it  is  evident  the  perimeter  of  the 
former  must  be  enlarged,  to  make  its  area  equal  to  the  area 
of  the  latter. 


PROPOSITION  II. 

80.  A  triangle  in  which  two  given  sides  make  a  right 
ANGLE,  has  a  greater  area  than  any  triangle  in  which  the  same 
sides  make  an  oblique  angle. 

If  BC,  BC,  and  BC"  (Fig.  27.)  be  equal,  and  if  BC  be 
perpendicular  to  AB ;  then  the  right  angled  triangle  ABC, 


ISOPERIMETRY.  63 

has  a  greater  area  than  the  acute  angled  triangle  ABC',  or 
the  oblique  angled  triangle  ABC". 

Let  PC  and  PC"  be  perpendicular  to  AP.  Then,  as  the 
three  triangles  have  the  sam^  base  AB,  their  areas  are  as 
their  heights  ;  that  is,  as  the  perpendiculars  BC,  PC',  and 
PC".  But  BC  is  equal  to  BC,  and  therefore  greater  than 
P'C  (Euc.  47,  1.)  BC  is  also  equal  to  BC",  and  therefore 
greater  than  PC". 


PROPOSITION    III. 

81.  If  all  the  sides  except  one  of  a  polygon  be  given^  the 
area  will  be  the  greatest^  when  the  given  sides  are  so  disposed^ 
that  the  figure  may  be  inscribed  in  a  semicircle,  of  which 
the  undetfrmined  side  is  the  diameter. 

If  the  sides  AB,  BC,  CD,  DE,  (Fig.  28.)  be  given,  and  if 
their  position  be  such  that  the  area,  included  between  these 
and  another  side  whose  length  is  not  determined,  is  a  maxi- 
mum ;  the  figure  maj  be  inscribed  in  a  semicircle,  of  which 
the  undetermined  side  AE  is  the  diameter. 

Draw  the  lines  AD,  AC,  EB,  EC.  Py  varying  the  angle 
at  D,  the  triangle  ADE  may  be  enlarged  or  diminished, 
without  affecting  the  area  of  the  other  parts  of  the  figure. 
The  whole  area,  therefore,  cannot  be  a  maximum^  unless 
this  triangle  be  a  maximum,  while  the  sides  AD  and  ED  are 
given.  But  if  the  triangle  ADE  be  a  jnaximumy  under  these 
conditions,  the  angle  ADE  is  a  right  single  ;  (Art.  80.)  and 
therefore  the  point  D  is  in  the  circumference  of  a  circle,  of 
which  AE  is  the  diameter.  (Euc.  31,  3.)  In  the  same  man- 
ner it  may  be  proved,  that  the  angles  ACE  and  ABE  arc  right 
angles,  and  therefore  that  the  points  C  and  B  are  in  the  cir- 
cumference of  the  same  circle. 

The  term  polygon  is  used  in  this  section  to  include  iriaii- 
glesj  and  four-sided  figures,  as  well  as  other  right-lined 
figures. 

82.  The  area  of  a  polygon,  inscribed  in  a  semi-circle,  in 
the  manner  stated  above,  will  not  be  altered  by  varying  the 
order  of  the  given  sides. 

The  sides  AB,  BC,  CD,  DE,  (Fig.  28.)  are  the  chords  of 
so  many  arcs.  The  sum  of  these  arcs,  in  whatever  order 
they  are  arranged,  will  evidently  be  equal  to  the  semicir- 
ciimference.     And  the  segments  between  the  given  sides  and 


64  MENSURATION. 

the  arcs  will  be  the  same,  in  whatever  part  of  the  circle  they 
are  situated.  But  the  area  of  the  polygon  is  equal  to  the 
area  of  the  semicircle,  diminished  by  the  sum  of  these  seg- 
ments. 

83.  If  a  polygon,  of  which  all  the  sides  except  one  are 
given,  be  inscribed  in  a  semicircle  whose  diameter  is  the  un- 
determined side ;  a  polygon  having  the  same  given  sides, 
cannot  be  inscribed  in  any  other  semicircle  which  is  either 
greater  or  less  than  this,  and  whose  diameter  is  the  undeter- 
mined side. 

The  given  sides  AB,  BC,  CD,  DE,  (Fig.  28.)  are  the 
chords  of  arcs  whose  sum  is  180  degrees.  But  in  a  larger 
circle,  each  would  be  the  chord  of  a  less  number  of  degrees, 
and  therefore  the  sum  of  the  arcs  would  be  less  than  180^  : 
and  in  a  smaller  circle,  each  would  be  the  chord  of  a  greater 
number  of  degrees,  and  the  sum  of  the  arcs  would  be  greater 
than  J80^ 


PROPOSITION  IV. 

84.  A  polygon  inscriijed  in  a  circle  has  a  greater  area, 
than  any  polygon  of  equal  perimeter,  and  the  same  number 
oj"  sides,  which  cannot  be  inscribed  in  a  circle. 

If  in  the  circle  ACHF,  (Fig.  30.)  there  be  inscribed  a  po- 
lygon ABCDEFG  ;  and  if  another  polygon  abcdefg  (Fig. 
31.)  be  formed  of  sides  which  are  the  same  in  number  and 
length,  but  which  are  so  disposed,  that  the  figure  cannot  be 
inscribed  in  a  circle  ;  the  area  of  the  former  polygon  is  greater 
than  that  of  the  latter. 

Draw  the  diameter  AH,  and  the  chords  DH  and  EH. 
Upon  de  make  the  triangle  deh  equal  and  similar  to  DEH, 
and  join  ah.  The  line  ah  divides  the  figure  abcdhefg  into  two 
parts,  of  which  one  at  least  cannot,  by  supposition,  be  inscri- 
bed in  a  semicircle  of  which  the  diameter  is  AH,  nor  in  any 
other  semicircle  of  which  the  diameter  is  the  undetermined 
side.  (Art.  83.)  It  is  therefore  less  than  the  corresponding 
part  of  the  figure  ABCDHEFG.  (Art.  81.)  And  the  other 
part  of  abcdhefg  is  not  greater  than  the  corresponding  part  of 
ABCDHEFG.  Therefore  the  whole  figure  ABCDHEFG  is 
greater  than  the  whole  figure  abcdhefg.  If  from  these  there 
be  taken  the  equal  triangles  DEH  and  deh,  there  will  remain 
the  polygon  ABCDEFG  greater  than  the  polygon  abcdefg. 


ISOPERIMETRY.  So 

85.  A  polygon  of  which  all  the  sides  are  given  in  number 
and  length,  can  not  be  inscribed  in  circles  of  different  diam- 
eters. (Art.  83,)  And  the  area  of  the  polygon  will  not  be 
altered,  by  changing  ti^e  order  of  the  sides.  (Art.  82.) 


PROPOSITION    V. 

86.    When  a  polygon  has  a  greater  area  than  any  other,  of 
the  same  number  of  sides,  and  of  equal  perimeter^  the  sides  are 

EQUAL. 

The  polygon  ABCDF  (Fig.  29.)  cannot  be  a  maximum^ 
among  all  polygons  of  the  same  number  of  sides,  and  of 
equal  perimeters  unless  it  be  equilateral.  For  if  any  two  of 
the  sides,  as  CD  and  FD,  are  unequal,  let  CH  and  FH  be 
equal,  and  their  sum  the  same  as  the  sum  of  CD  and  FD. 
The  isosceles  triangle  CHF  is  greater  than  the  scalene  trian- 
gle CDF  (Art.  79.)  ;  and  therefore  the  polygon  ABCHF  is 
greater  than  the  polygon  ABCDF  ;  so  that  the  latter  is  not  a 
maximum* 


PROPOSITION     VI. 

07.  Jl  REGULAR  POLYGON  has  fl  greater  area  than  any  other 
polygon  of  equal  perimeter,  and  of  the  same  number  of  sides. 

For,  by  the  preceding  article,  the  polygon  which  is  a  max- 
imum among  others  of  equal  perimeters,  and  the  same  num- 
ber of  sides  is  equilateral,  and  by  art.  84,  it  may  be  inscribed 
in  a  circle.  But  if  a  polygon  inscribed  in  a  circle  is  equilat- 
eral, as  ABDFGH  (Fig.  7.)  it  is  also  equiangular.  For  the 
sides  of  the  polygon  are  the  bases  of  so  many  isosceles  trian- 
gles, whose  common  vertex  is  the  centre  C.  The  angles  at 
these  bases  are  all  equal  ;  and  two  of  them,  as  AHC  and 
GHC,  are  equal  to  AHG  one  of  the  angles  of  the  polygon. 
The  polygon,  then,  being  equiangular,  as  well  as  equilateral, 
is  2i  regular  polygon.  (Art.  1.  Def.  2.) 

Thus  an  equilateral  triangle  has  a  greater  area,  than  any 
other  triangle  of  equal  perimeter.  And  a  square  has  a  greater 
area,  than  any  other  four-sided  figure  of  equal  perimeter. 

10 


ee  MENSURATION. 

Cor.  A  regular  polygon  has  a  less  perimeter  than  any  other 
polygon  of  equal  area,  and  the  same  number  of  sides. 

For  if,  with  a  given  perimeter,  the  regular  polygon  is  greater 
than  one  which  is  not  regular;  it  is  evident  the  perime- 
ter of  the  former  must  be  diminished,  to  make  its  area  equal 
to  that  of  the  latter. 


PROPOSITION    VII. 

88.  If  a  polygon  be  described  about  a.  circle,  the  areai 

of  the  two  figures  are  as  their  perimeters* 

Let  ST  (Fig.  32.)  be  one  of  the  sides  of  a  polygon,  either 
regular  or  not,  which  is  described  about  the  circle  LNR. 
Join  OS  and  OT,  and  to  the  point  of  contact  M  draw  the 
radius  OM,  which  will  be  perpendicular  to  ST.  (Euc.  18,  3.) 
The  triangle  OST  is  equal  to  half  the  base  ST  raultiphed 
into  the  radius  OM.  (Art.  8.)  And  if  lines  be  drawn,  in  the 
same  manner,  from  the  centre  of  the  circle,  to  the  extremi- 
tfes  of  the  several  sides  of  the  circumscribed  polygon,  each 
of  the  triangles  thus  formed  will  be  equal  to  half  its  base  mul- 
tiplied into  the  radius  of  the  circle.  Therefore  the  area  of 
the  whole  polygon  is  equal  to  half  its  perimeter  multiplied 
into  the  radius  :  and  the  area  of  the  circle  is  equal  to  half  its 
circumference  multiplied  into  the  radius.  (Art.  30.)  So  that 
the  two  areas  are  to  each  other  as  their  perimeters. 

Cor.  I.  If  different  polygons  are  described  about  the  same 
circle,  their  areas  are  to  each  other  as  their  perimeters.  For 
the  area  of  each  is  equal  to  half  its  perimeter,  multiplied  into 
the  radius  of  the  inscribed  circle. 

Cor.  2.  The  tangent  of  an  arc  is  always  greater  than  the 
arc  itself.  The  triangle  OMT  (Fig.  32.)  is  to  OMN,  as  MT 
to  MN.  But  OMT  is  greater  than  OMN,  because  the  for- 
mer includes  the  latter*  Therefore  the  tangent  MT  is  greater 
than  the  arc  MN. 


ISOPERIMETRY.  ^7 


PROFOSITION    VIII, 

89.  Jl  CIRCLE  has  a  greater  area  than  any  polygon  of  equal 
perimeter. 

If  a  circle  and  a  regular  polygon  have  the  same  centre, 
and  equal  perimeters;  each  of  the  sides  of  the  polygon  must 
fall  partly  within  the  circle.  For  the  area  of  a  circumscribing 
polygon  is  greater  than  the  area  of  the  circle,  as  the  one  in- 
cludes the  other:  and  therefore,  by  the  preceding  article,  the 
perimeter  of  the  former  is  greater  than  that  of  the  latter. 

Let  AD  then  (Fig.  32.)  be  one  si<le  of  a  regular  polygon, 
whose  perimeter  is  equal  to  the  circumference  of  the  circle 
RLN.  As  this  falls  partly  within  the  circle,  the  perpendicu- 
lar OP  is  less  than  the  radius  OR.  But  the  area  of  the  poly- 
gon is  equal  to  half  its  perimeter  multiplied  into  this  perpen- 
dicular (Art.  15.);  and  the  area  of  the  circle  is  equal  to  half 
its  circumference  multiplied  into  the  radius.  (Art.  30.)  The 
circle  then  is  greater  than  the  given  regular  polygon ;  and 
therefore  greater  than  any  other  polygon  of  equal  perimeter. 
(Art.  87.) 

Cor.  1 .  A  circle  has  a  less  perimeter ^  than  any  polygon  of 
equal  area. 

Cor.  2  Among  regular  polygons  of  a  given  perimeter, 
that  which  has  the  greatest  number  of  sides,  has  also  the  greats 
est  area.  For  the  greater  the  number  of  sides,  the  more 
nearly  does  the  perimeter  of  the  polygon  approach  to  a  coin- 
/cidence  with  the  circumference  of  a  circle.* 


PBOPOSITION    IX. 

90.  A  right  PRISM  whose  hises  are  regular  polygons,  has  a  less 
surface  than  any  other  riglt  prism  of  the  same  solidity^  the  same 
altitude,  and  the  same  nvtnber  of  sides. 

If  the  altitude  of  a  prism  is  given,  the  area  of  the  base  is 

as  the  solidity  (Art.  43.) ;  and  if  the  number  of  sides  is  also 
given,  the  perimeter  is  a  minimum  when  the  base  is  a  regular 

*  For  a  rigorous  demoDstratioD  of  this,  see  Legendre's  Geometry,  Appen- 
dix to  Bookiv. 


68  MENSURATION. 

polygon.  (Art.  87.  Cor.)  But  the  lateral  surface  is  as  the 
perimeter  (Art.  47.)  Of  two  right  prisms,  then,  which  have 
the  same  altitude,  the  same  solidity,  and  the  same  number  of 
sides,  that  whose  bases  are  regular  polygons  has  the  least 
lateral  surface,  while  the  areas  of  the  ends  are  equal. 

Cor.  A  right  prism  whose  bases  are  regular  polygons  has 
a  greater  solidity,  than  any  other  right  prism  of  the  same  sur- 
face, the  same  altitude,  and  the  same  number  of  sides. 


PROPOSITION    X. 

91.  .^  right  CYLINDER  has  a  less  surface^  than  any  right  prism  qf 
the  same  altitude  and  solidity. 

For  if  the  prism  and  cylinder  have  the  same  altitude  and 
solidity,  the  areas  of  their  bases  are  equal.  (Art.  64.)  But 
the  perimeter  of  the  cylinder  is  less,  than  that  of  the  prism 
(Art.  89.  Cor-  1.);  and  therefore  its  lateral  surface  is  less^ 
while  the  areas  of  the  ends  are  equal. 

Cor.  A  right  cylinder  has  a  greater  solidity^  than  any  right 
prism  of  the  same  altitude  and  surface. 


PROPOSITION    XI. 

92.  A  CUBE  has  a  less  surface  than  any  other  right  parallelepiped 
of  the  same  solidity. 

A  parallelopiped  is  a  prism,  any  one  of  whose  faces  may 
be  considered  a  base.  (Art.  41.  Def.  I.  and  V.)  If  these 
are  not  all  squares,  let  one  which  is  not  a  square  be  taken  for 
a  base.  The  perimeter  of  this  may  ho.  diminished,  without' 
altering  its  area  (Art.  87.  Cor.)  ;  and  therefore  the  surface  of 
the  solid  may  be  diminished,  without  altering  its  altitude  or 
solidity.  (Art.  43,  47.)  The  same  a«y  be  proved  of  each 
of  the  other  faces  which  are  not  squares.  The  surface  is 
therefore  a  minimum,  when  all  the  faces  are  squares,  that  is, 
when  the  solid  is  a  cube. 

Cor.  A  cube  has  a  greater  solidity  thsin  any  other  right 
parallelopiped  of  the  same  surface. 


IPSOPERIMETRY.  63 


PROPOSITION  XII. 


93.  AcvBE  has  a  greater  solidity^  than  any  other  right  parallelopi- 
ved^  the  ^um  of  whose  lengthy  breadth^  and  depth  is  equal  to  the  svm 
,of  the  corresponding  dimensions  of  the  cube. 

The  solidity  is  equal  to  the  product  of  the  length,  breadth, 
and  depth.  If  the  length  and  breadth  are  unequal,  the  solid- 
ity may  be  increased,  without  altering  the  sum  of  the  three 
dimensions.  For  the  product  of  two  factors  whose  sum  is 
given,  is  the  greatest  when  the  factors  are  equal.  (Euc.  27.  6.) 
In  the  same  manner,  if  the  breadth  and  depth  are  unequal, 
the  solidity  may  be  increased,  without  altering  the  sura  ot  the 
three  dimensions.  Therefore,  the  solid  can  not  be  a  maxi- 
mum, unless  its  length,  breadth,  and  depth  are  equal. 


PROPOSITION    XIII. 

94,  If  a  PRISM  BE  DESCRIBED  ABOUT  A  CYLINDER,  the  Capacities  of 
the  two  solids  are  a,s  their  surfaces. 

The  capacities  of  the  solids  are  as  the  areas  of  their  bases, 
that  is.  as  the  ^erimefer*  of  their  bases.  (Art.  88.)  But  the 
lateral  surfaces  are  also  as  the  perimeters  of  the  bases.  There- 
fore  the  whole  surfaces  are  as  the  solidities. 

Cor.  The  capacities  of  different  prisms,  described  about 
the  same  right  cylinder,  are  to  each  other  as  their  surfaces. 


PROPOSITION    XIV. 

95.  A  right  cylinder  whose  height  is  equal  to  the  diameter  of^ 
ITS  BASE  has  a  greater  solidity  than  any  other  right  cylinder  of  equal 
surface. 

Let  C  be  a  right  cylinder  whose  height  is  equal  to  the  di- 
ameter of  its  base  ;  and  C  another  right  cylinder  having  the 
same  surface,  but  a  different  altitude.  Tf  a  square  prism  P 
be  described  about  the  former,  it  will  be  a  cube.  But  a 
square  prism  P'  described  about  the  latter  will  not  be  a  cube. 


70  MENSURATION. 

Then  the  surfaces  of  C  and  P  are  as  their  bases  (Arts'. 
47  and  88.) ;  which  are  as  the  bases  of  C  and  P'  (Sup,  Euc. 
7,  1.)  ;  so  that, 

surjC :  surfP : :  haseC  :  hase? : :  hastC  :  hase?' ::  surfC' :  awr/P' 
But  the  surface  of  C  is,  by  supposition,  equal  to  the  sur- 
face of  C.     Therefore,  (Alg.  395.)  the  surface  of  P  is  equal 
to  the  surface  of  P'.     And  by  the  preceeding  article, 
fiolidP  :  solidC'.'.surfF  :  surfC  ::  surfP' :  surfC'iisolidP':  solidC* 

But  the  solidity  of  P  is  greater  than  that  of  P'.  (Art.  92. 
Cor.)     Therefore  the  solidity  of  C  is  greater  than  that  of  C'. 

Schol.  A  right  cylinder  whose  height  is  equal  to  the  diam- 
eter of  its  base,  is  that  which  circumscribes  a  sphere.  It  is 
also  called  Archimedes''  cylinder  ;  as  he  discovered  the  ratio 
of  a  sphere  to  its  circumscribing  cylinder ;  and  these  are  the 
igures  which  were  put  upon  his  tomb. 

Cor.  Archimedes'  cylinder  has  a  less  surface^  than  any 
other  right  cylinder  of  the  same  capacity. 


rROPOSlTION    XV, 

96.  If  a  SPHERE  BE  CIRCUMSCRIBED  by  a  solid  bounded  by  plane  sw- 
faces  ;  the  capacities  of  the  two  solids  are  cw  their  surfaces. 

If  planes  be  supposed  to  be  drawn  from  the  centre  of  the 
sphere,  to  each  of  the  edges  of  the  circumscribing  solid,  they 
will  divide  it  into  as  many  pyramids  as  the  solid  has  faces. 
The  base  of  each  pyramid  will  be  one  of  the  faces;  and  the 
height  will  be  the  radius  of  the  sphere.  The  capacity  of  the 
pyramid  will  be  equal,  therefore,  to  its  base  multiplied  into  | 
of  the  radius  (Art.  48  );  and  the  capacity  of  the  whole  cir- 
cumscribing solid,  must  be  equal  to  its  whole  surface  multi- 
plied into  1  of  the  radius.  But  the  capacity  of  the  sphere  is 
also  equal  to  its  surface  multiplied  into  ^  of  its  radius, 
(Art.  70.) 

Cor.  The  capacities  of  different  solids  circumscribing  the 
same  sphere,  are  as  their  surfaces. 


ISOPERIMETRY.  71 


PROPOSITION    XVI, 


97.  Ji  SPHERE  has  a  greater  solidity,  than  any  regular  polyedron  of 
e(iual  surface. 

If  a  sphere  and  a  regular  polyedron  have  the  same  centre, 
and  equal  surfaces;  each  of  the  faces  of  the  polyedron  must 
fall  partly  within  the  sphere.  For  the  solidity  of  a  circum- 
scribing solid  is  greater  than  the  solidity  of  the  sphere,  as  the 
one  includes  the  other:  and  therefore,  by  the  preceding  arti- 
cle, the  surface  of  the  former  is  greater  than  that  of  the  latter. 

But  if  the  faces  of  the  polyedron  fall  partly  within  the 
sphere,  their  perpendiculac  distance  from  the  centre  must  be 
less  than  the  radius.  And  therefore,  if  the  surface  of  the 
polyedron  be  only  equal  to  that  of  the  sphere,  its  solidity 
must  be  less.  For  the  solidity  of  the  polyedron  is  equal  to 
its  surface  multiplied  into  \  of  the  distance  from  the  centre. 
(Art.  59.)  And  the  solidity  of  the  sphere  is  equal  to  its  sur- 
face multiplied  into  \  of  the  radius. 

Cor.  A  sphere  has  a  less  surface^  than  any  regular  polye- 
dron of  the  same  capacity. 

For  ©tber  casts  of  Isoperimetry,  see  Fluxions. 


APPENDIX.— PART  I. 


Containing  rules,  without  demonstrations,  for  the  mensuration 
of  the  Come  Sections^  and  other  figures  not  treated  of  in 
the  Elements  of  Euclid,* 


PROBLEM    I. 

To  find  the  area  of  an  ellipse. 

101.  Multiply  the  product  of  the  transverse  and  conjugate 
axes  into  .7854. 

Ex.  What  is  tlie  area  of  an  ellipse  whose  transverse  axis  is 
36  feet,  and  conjugate  28  ?  Ans.  791.68  feet. 

PROBLEM    II. 

To  find  the  area  of  a  segment  of  an  ellipse,  cut  off  by  a  line 
perpendicular  to  either  axis. 

102.  If  either  axis  of  an  ellipse  be  nnade  the  diameter  of  a 
circle ;  and  if  a  line  perpendicular  to  this  axis  cut  off  a  seg- 
ment from  the  ellipse,  and  from  the  circle ; 

The  diameter  of  the  circle  is,  to  the  other  axis  of  the  ellipse  : 
As  the  circular  segment,  to  the  elliptic  segment. 

*  For  demonstrations  of  these  rules,  see  Conic  Sections,  Spherical  Trig©- 
ncHuetry,  and  Fluxions,  or  Hutton's  Mensuration* 


APPENDIX.  73 

Kx.  What  is  the  area  of  a  segment  cut  oti'from  an  ellipse 
whose  transverse  axis  is  415  feet,  and  conjugate  33*2 ;  if  the 
height  of  the  segment  is  06  feet,  and  its  base  is  perpendicu- 
lar to  the  transverse  ffxis  ? 

The  circular  segment  is  23G0O  feet. 

And  the  elliptic  segment  18944. 


PROBLEM    III. 

To  find  the  area  of  a  conic  parabola. 

103.  Multiply  the  base  by  f  of  the  height. 

Ex.  If  the  base  of  a  parabola  is  26  inches,  and  the 
height  9  feet;  what  is  the  area  ?  Ans.  13  feet. 

PROBLEM    IV. 

To  find  the  area  of  a  frustum  of  a  parabola^  cut  off  by  a  line  par- 
allel to  the  base. 

104.  Divide  the  difference  of  the  cubes  of  the  diameters 
of  the  two  ends,  by  the  ditTerence  of  their  squares  ;  and  mul- 
tiply the  quotient  by  f  of  the  perpendicular  height. 

Ex.  What  is  the  area  of  a  parabolic  frustum,  whose  height 
is  12  feet,  and  the  diameters  of  its  ends  20  and  12  feet? 

Ans.  196  feet. 

problem  v. 
To  find  the  area  of  a  conic  hvperhola. 

105.  Multiply  the  base  by  |  of  the  height;  and  correct 
ihe  product  by  subtracting  from  it  the  series 

-*''><  (rf:5+ 3X7+^+ fxrr  +^^-) 

C6  =  the  base  or  double  ordinate, 
Fn  which    </?=ihe  height  or  abscissa, 

(2:  =  the  height  divided  by  the  sum  of 
the  height  and  transverse  axis. 
11 


74  MENSURATION. 

The  series  converges  so  rapidly,  that  a  few  of  the  first 
terms  will  generally  give  the  correction  with  sufficient  exact- 
ness. This  correction  is  the  difference  between  the  hyper- 
bola, and  a  parabola  of  the  same  base  and  height. 

Ex.  If  the  bas»of  a  hyperbola  be  24  feet,  the  height  10 
and  the  transverse  axis  30  ;  what  is  the  area  ? 
The  base  X  |  the  height  is  1 60. 

The  first  term  of  the  series  is  0.016666 

The  second  0.000592 

The  third  0.000049 

The  fourth  0.000006 


Their  suoi  0.017313 


This  into  2bh'\8  8.31 


And  the  area  corrected  is  151.69 


PROBLEM  VI. 


To  find  the  area  of  a  spherical  triangle  formed  by  three  arcs  of 
great  circles  of  a  sphere. 


106.  As  0  right  angles  or  720°, 

To  the  excess  of  the  3  given  angles  above  1 80°  ; 
So  is  the  whole  surface  of  the  sphere, 
To  the  area  of  the  spherical  triangle. 

Ex.  What  is  the  area  of  a  spherical  triangle,  on  a  sphere 
whose  diameter  is  30  feet,  if  the  angles  are  130°,  102°,  and 
68'»?  Ans.  471.24  feet. 

PROBLEM  VII. 

To  find  the  area  of  a  spherical  polygon  formed  by  arcs  of  great 

circles. 

107.  As  0  right  angles,  or  720°, 

To  the  excess  of  all  the  given  angles  above  the  pro- 
duct of  the  number  of  angles  -  2  into  1 80*  •, 
So  is  the  whole  surface  of  the  sphere, 
To  the  area  of  the  spherical  polygon. 


APPENDIX.  76 


£x.  What  is  the  area  of  a  spherical  polygon  of  seven 
sides,  on  a  sphere  whose  diameter  is  17  inches ;  if  the  sum 
of  all  the  angles  is  1080°  ?  Ans.  227  inches. 


PROBLEM    VIIl. 

To  find  the  lunar  surface  included  between  two  great  circles  cf 

a  sphere. 

108.  As  360°,  to  the  angle  made  by  the  given  circles ; 

So  is  the  whole  surface  of  the  sphere,  to  the  surface 
between  the  circles. 
Or, 
The  lunar   surface  is   equal  to  the  breadth  of  the  middle 
part  of  it,  multiplied  into  the  diameter  of  the  sphere. 

£x.  If  the  earth  be  7930  miles  in  diameter,  what  is  the 
surface  of  that  part  of  it  which  is  included  between  the  65lh 
and  83d  degree  of  longitude  f 

Ans.  9,878,000  square  miles. 


PROBLEM    IX. 

To  find  the  solidity  of  a  spheroid,  formed  by  the  revolution  of  an 
ellipse  about  either  axis. 

109.  Multiply  the  product  of  the  fixed  axis  and  the  square 
of  the  revolving  axis,  into  .5236. 

Ex.  1.  What  is  the  solidity  of  an  oblong  spheroid,  whose 
longest  and  shortest  diameters  are  40  and  30  feet  ? 

Ans.  40x5o^  X.5236  =  16850  feet. 

2.  If  the  earth  be  an  oblate  spheroid,  whose  polar  and 
equatorial  diameters  are  7930  and  7960  miJ«s;  what  is  its 
solidity  f  Ans.  263,000,000,000  miles. 


MENSURATION. 


PROBLEM    X. 


To  Jind  the  solidity  of  the  middle  frustum  of  a  spheroid^  included 
between  two  planes  which  are  perpendicular  to  the  axis,  and 
equally  distant  from  the  centre. 

110.  Add  together  the  square  of  the  diameter  of  one  end, 
and  twice  the  square  of  the  middle  diameter;  multiply  the 
sum  by  -J  of  the  height,  and  the  product  by  .7834. 

If  D  and  d  =  the  two  diameters,  and  h  =  the  height  ; 
The  solidity  =(2D-  +rf2 )  x  ^A  X  .7854. 

Ex.  If  the  diameter  of  one  end  of  a  middle  frustum  of 
a  spheroid  be  8  inches,  the  middle  diameter  10,  and  the 
height  30,  what  is  the  solidity  ^ 

Ans.  2073.4  inches. 

Cor.  Half  the  middle  frustum  is  equal  to  a  frustum  of 
which  one  of  the  ends  passes  through  the  centre. 

If  then  D  and  d=xhe  diam'rs  of  the  twoends,  and  A=the  hei't, 
The  solidity  =(2D^+(/2)  x^Ax  .7854. 


PROBLEM    XI. 

To  Jind  the  solidity  of  a  paraboloid. 

111.  Multiply  the  area  of  the  base  by  half  the  height. 

Ex    If  the  diameter  of  the  base  of  a  paraboloid  be  12  feet, 
and  the  height  22  feet,  what  is  the  solidity  ? 

Ans.  1243  feet. 

PROBLEM    XII. 

To  find  the  solidity  of  a  frustum  of  a  paraboloid, 

112.  Multiply  the  sum  of  the  areas  of  the  two  ends  by  half 
their  distance. 


APPENDIX.  77 

Ex.  If  the  diameter  of  one  end  of  a  frustum  of  a  parabo- 
loid be  8  feet,  the  diameter  of  the  other  end  6  feet,  and  the 
length  21  feet ;  what  is  the  solidity  ? 

Ans.  942i  feet. 

Cor.  If  a  cask  be  in  the  form  of  fzco  equal  frustums  of  a 
paraboloid ;  and 

If  D  =  the  middle diam.  r/=the  end  diam.  and  h  =  ihe  length; 
The  solidity  =  (D2-f-f/3)xi/iX. 7854. 


PaOBLCJI  XIII. 

To  Jind  the  solidity  of  a  hyperboloid,  produced  by  the  revolu- 
tion of-a  hyperbolii  on  its  axis. 

113.  Add  together  the  square  of  the  radius  of  the  base, 
and  the  square  of  the  diameter  of  a  section  which  is  equally 
distant  from  the  base  and  the  vertex  ;  multiply  the  sura  by 
the  height,  and  the  product  by  .^236. 

If  R  =  the  radius  of  the  base,  D=the  middle  diameter,  and 
h=ihe  height ; 

The  solidiiy  =  (R=+D-^)xAx.5236. 

Ex.  If  the  diameter  of  the  base  of  a  hyperboloid  be  24, 
the  square  of  the  middle  diameter  252,  and  the  height  10, 
wliat  is  the  solidity  ?  Ans.  2073.4. 


PROBLEM  XIV. 

Ta  find  the  solidity  of  a  frustum  of  a  hyperboloid. 

114.  Add  together  the  squares  of  the  radii  of  the  two  ends, 
and  the  square  of  the  middle  diameter;  multiply  the  sum  by 
the  height,  and  the  product  by  .5236. 

If  R  and  r=the  two  radii,  D  =  the  middle  diameter,   and 
h  =  the  height; 
The  solidity =(R- 4-r-  +D-^ )xhX .5236. 

Ex.  If  the  diameter  of  one  end  of  a  frustum  of  a  hyperbo- 
loid be  32,  the  diameter  of  the  other  end  24,  the  square  of 
the  middle  diameter  793|,  and  the  length  20,  what  is  the 
solidity?  Ans.  12^09.3. 


78  MENSURATION. 


PROBLEM    X<r> 

To  find  the  solidity  of  a  circular  spindle,  produced  by  the  revo- 
lution of  a  circular  segment  about  its  base  or  chord  as  an 

axis. 

115.  From  J  of  the  cube  of  half  the  axis,  subtract  the 
product  of  the  central  distance  into  half  the  revolving  cir- 
cular segment,  and  multiply  the  remainder  bv  four  times 
^.14159. 

If  a=the  area  of  the  revolving  circular  segment, 
/=:half  the  length  or  axis  of  the  spindle, 
e=the  distance  of  this  axis  from  the  centre  of  the 
circle  to  which  the  revolving  segment  belongs  ; 

The  solidity=(|/3-iac)X4X3.14159. 

Ex.  Let  a  circular  spindle  be  produced  by  the  revolution 
of  the  segment  ABO  (Fig.  9.)  about  AB.  If  the  axis  AB  be 
140,  and  OP  half  the  middle  diameter  of  the  spindle  be 
38.4  ;  what  is  the  solidity  ? 

The  area  of  the  revolving  segment  is        3791 
The  central  distance  PC  44.6 

The  solidity  of  the  spindle  374402 


problem  XVI. 

To  find  the  solidity  of  the  middle  frustum  of  a  circular  spindlt. 

1 16.  From  the  square  of  half  the  axis  of  the  whole  spindle, 
subtract  ^  of  the  square  of  half  the  length  of  the  frustum ; 
multiply  the  remainder  by  this  half  length ;  from  the  product 
subtract  the  product  of  the  revolving  area  into  the  central 
distance;  and  multiply  the  remainder  by  twice  3.14169. 

If  L=half  the  length  or  axis  of  the  whole  spindle, 
/=half  the  length  of  the  middle  frustum, 
c=the  distance  of  the  axis  from  the  centre  of  the  circle, 
rt=the  area  of  the  figure  which,  by  revolving,  pro 

duces  the  frustum  ; 
The  solidity =(l7^rp7x/--ac)X2X 3.1 41 59. 


APPENDIX.  79 

Ex.  If  the  diameter  of  each  end  of  a  frustum  of  a  circular 
spindle  be  21.6,  the  middle  diameter  60,  and  the  length  70; 
what  is  the  solidity  i* 

The  length  of  the  whole  spindle  is  79.75 

The  central  distance  11.5 

The  revolving  area  1703.8 

The  solidity  136751.5 


PROBLEM    XVII. 

To  find  the  solidity  of  a  parabolic  spindle,  produced  by  the  revo- 
lution of  a  parabola  about  a  double  ordinate  or  base. 

117.  Multiply  the  square  of  the  middle  diameter  by  f*^  of 
the  axis,  and  the  product  by  .7854. 

Ex.  If  the  axis  of  a  parabolic  spindle  be  30,  and  the  mid- 
dle diameter  17,  what  is  the  solidity  ? 

Ans.  3631.7. 


PROBLEM    XVIII. 

To  find  the  solidity  of  the  middle  frustum  of  a  parabolic  spindle. 

118.  Add  together  the  square  of  the  end  diameter,  and 
twice  the  square  of  the  middle  diameter,  from  the  sum  sub- 
tract I  of  the  square  of  the  difference  of  the  diameters;  and 
multiply  the  remainder  by  |  of  the  length,  and  the  product 
by  .7854. 

If  D  and  d  =  the  two  diameters,  and  /  =  the  length; 
The  solidity  =(2D2+rf»-|(D~rf)2)Xi/X.7854. 

Ex.  If  the  end  diameters  of  a  frustum  of  a  parabolic  spin- 
dle be  each  ]2inches,  the  middle  diameter  16,  and  the  length 
30  ;  what  is  the  solidity  ?  Ans.  5102  inches. 


APPENDIX.— PART  11. 


GAUGING  OF  CASKS. 


At  11Q  ^  AUGING  is  a  practical  art,  whicli  does  not 
*  ^^  admit  of  being  treated  in  a  very  scientific 
manner.  Casks  are  not  commonly  constructed  in  exact  con- 
formity with  any  regular  mathematical  figure.  By  most  wri- 
ters on  the  subject,  however,  they  are  considered  as  nearly 
coinciding  with  one  of  the  following  forms  ; 

'  i  The    middle   frustum  <    ^  u  r  '      •  ji 

2.5  ^  of  a  parabolic  spmdie. 

3. )  rp  1    r      .         (  of  a  paraboloid, 

-    >  Iwo   equal    irustums  <    c     ^  ' 

4.  )  ^  ( ot  a  cone. 

The  second  of  these  varieties  agrees  more  nearly,  than  any 
of  the  others,  with  the  forms  of  casks,  as  they  are  commonly 
made.  The  first  is  too  much  curved,  the  third  too  little,  and 
the  fourth  not  at  all,  from  the  head  to  the  bung. 

120.  Rules  have  already  been  given,  for  finding  the  capa- 
city of  each  of  the  four  varieties  of  casks.  (Arts.  68,  110, 
112,  118.)  As  the  dimensions  are  taken  in  Inches,  these 
rules  will  give  the  contents  in  cubic  inches.  To  abridge  the 
computation,  and  adapt  it  to  the  particular  measures  used  in 
gauging,  the  factor  .7854  is  divided  by  282  or  231 ;  and  the 
quotient  is  used  instead  of  .7854,  for  finding  the  capacity  in 
ale  gallons  or  wine  gallons. 
.7854 
Now-2g2-'=.002785,or  .0028  nearly; 

.    ,.■7854 

And  ^^=.0034. 

If  then  .0028  and  ,0034  be  substituted  for  .7854,  in  the 
rules  referred  to  above ;  the  contents  of  the  cask  will  be  giv- 
en in  ale  gallons  and  wine  gallons.  These  numbers  are  to 
each  other  nearly  as  9  to  11. 


GAUGING.  gi 


PROBLEM    I. 


To  calculate  iht  mmtcnis  of  a  cask^  in  the  form  of  the  middle  frus- 
tum  of  a  SPHEROID. 

121.  Add  together  the  square  of  the  head  diameter,  and 
twice  the  square  of  the  bung  diameter ;  multiply  the  sum  by 
I  of  the  length,  and  the  product  by  .0028  for  ale  gallons,  or 
by  0034  for  wine  gallons. 

If  D  and  </=the  two  diameters,  and  Z=the  length; 
The  capacity  in  inches  =  (2D=4-rf2)xi/x. 7654.  (Art.  110.) 
And  by  substituting  .0023  or  .0034  for  .7854,  we  have  the 
capacity  in  ale  gallons  or  wine  gallons. 

Ex.  What  is  the  capacity  of  a  cask  of  the  first  form, 
whose  length  is  30  inches,  its  head  diameter  18,  and  its  bung 
diameter  24.^ 

Ads.  41.3  ale  gallons,  or  52.2  wine  gallons. 


PROBLEM    II. 

To  calculate  the  contents  of  a  cask,  in  the  form  of  the  middle  frus- 
tum of  a  PARABOLIC    SPINDLE. 

122.  Add  together  the  square  of  the  head  diameter,  and 
twice  the  square  of  the  bung  diameter,  and  from  the  sura 
subtract  f  of  the  square  of  the  difference  of  the  diameters  ; 
multiply  the  remainder  by  |  of  the  length,  and  the  product 
by  .0028  for  ale  gallons,  or  .0034  for  wine  gallons. 

The  capacity  in  inches  =(20^  4-d2-|(D-£?)2)  x^/x 
.7854.  (Art.  118.) 

Ex.  What  is  the  capacity  of  a  cask  of  the  second  form, 
whose  length  is  30  inches,  iis  head  diameter  18,  and  its  bung 
diameter  24 .'' 

Ans,  40.9  ale  gallons,  or  49.7  wine  gallons* 
12 


82  MENSURATION. 


PROBLEM    III. 


To  calculate  the  contents  of  a  cask^  in  the  form  of  two  equal  frus- 
tums of  a   FARABOLOID. 

123.  Add  together  the  square  of  the  head  diameter,  and 
the  square  of  the  bung  diameter;  muhiply  the  sum  by  half 
the  length,  and  the  product  by  .0028  for  ale  gallons,  or  .0034 
for  wine  gallons. 

The  capacity  in  iaches=(D2-f£;3)xiZx. 7854.  (Art. 
112.  Cor.) 

Ex.  What  is  the  capacity  of  a  cask  of  the  third  form, 
whose  dimensions  are,  as  before,  30,  18,  24? 

Ans.  37.8  ale  gallons,  or  45.9  wine  gallons. 


PROBLEM    IV. 

To  calculate  the  contents  of  a  cask^  in  thejorm  of  two  equal  frustums 

of  a    CONE. 

124.  Add  together  the  square  of  the  head  diameter,  the 
square  of  the  bung  diameter,  and  the  product  of  the  two  di- 
ameters; multiply  the  sum  by  i  of  the  length,  and  the  pre 
duct  by  .0028  for  ale  gallons,  or  0034  for  wine  gallons. 

The  capacity  in  inches=(D2  +rf2  ^Drf)  X  i/  X  .7854  (Art.68.) 

Ex.  What  is  the  capacity  of  a  cask  of  the  fourth  form, 
whose  length  is  30.  and  its  diameters  IS  and  24  ^ 

Ans.  37.3  ale  gallons,  or  45.3  wine  gallons. 

125.  The  preceding  rules,  though  correct  in  theory,  are 
not  very  well  adapted  to  practice,  as  they  suppose  the  form 
of  the  cask  to  h^.  known.  The  two  following  rules,  taken 
from  Button  s  Mensuration,  may  be  used  for  casks  oi  the 
usual  forms.  For  the  first,  three  dimensions  are  required,  the 
length,  the  head  diameter,  and  the  bung  diameter.  It  is  ev- 
ident that  no  allowance  is  made  by  this,  for  different  degrees 
of  curvature  from  the  head  to  the  bung.  If  the  cask  is  more 
or  less  curved  than  usual,  the  following  rule  is  to  be  prefer- 
red, for  which  four  dimensions  are  required,  the  head  and 


GAUGING.  83 

bung  diameters,  and  a  third  diameter  taken  in  the  middle 
between  the  bune  and  the  head.     For  the  drmonstf  ttiotj  of 
these  rules,  see  Hutton's  Mensuration,  Part  v.  Sec.  2.  Ch.  5. 
and  7. 


PROBLEM  V. 

To  calculate    the    contents    of  any  common  cask  from  three  di- 
mensions. 

12C.  Add  together 

25  times  the  square  of  the  head  diameter, 

39  times  tl^e  square  of  the  bung  diameter,  and 

26  times  th'   product  of  the  two  diameters  ; 
Multiply  the  sum  by  the  length,  divide  the  product  by  90, 

and  multiply  the  quotient  by  .0028  for  ale  gallons,  or  .0034 
for  wine  gallons. 

Thecapacityininches=(39D2  4-25c?2-f26DJ)  X— X.7854. 

Ex.  What  is  the  capacity  of  a   cask  whose  length  is  30 
inches,  the  head  diameter  18,  and  the  bung  diameter  24  ? 
Ans.  39  ale  gallons,  or  47|  wine  gallons. 


PROBLEM  VI. 

To  calculate  the  contents  of  a  cask  from  four  dimensions^  the 
length,  the  head  and  bung  diameters,  and  a  diameter  taken 
in  the  middle  between  the  bead  and  the  bung. 

127.  Add  together  the  squares  of  the  head  diameter,  of 
the  bung  diameter,  and  of  double  the  middle  diameter;  mul- 
tiply the  sum  by  }  of  the  length,  and  the  product  by  .0028 
for  ale  gallons,  or  .0034  for  wine  gallons. 

If  D=the  bung  diameter,  d=ihe  head  diameter,  m=the 
middle  diameter,  and  /=the  length  ; 

The  capacity  in  inches  =  (D2 -|-c?2 -f^^^)  X|/ X.7854. 

Ex.  What  is  the  capacity  of  a  cask,  whose  length  is  30 

inches,  the  head  diameter  18,  the  bung  diameter  24,  and  the 
1 " 

Ans.  41  ale  gallons,  or  49|  wine  gallons. 


(4      •  MENSURATION. 

128.  In  making  the  calculations  in  gauging,  according  to 
Ihe  preceding  rules,  the  multiplications  and  divisions  are  fre- 
quently performed  by  means  of  a  Sliding  Rule^  on  which 
are  placed  a  number  of  logarithmic  lines,  similar  to  those  on 
Gunter's  Scale.     See  '1  rigonom.  Sec.  vi.  and  Note  I.  p.  122. 

Another  instrument  commonly  used  in  gauging  is  the 
Diagonal  Rod.  By  this,  the  capacity  of  a  cask  is  very  ^ex- 
peditiously found,  from  a  single  dimension,  the  distance  from 
the  bung  to  the  intersection  of  the  opposite  stave  with  the 
head.  The  measure  is  taken  by  extending  the  rod  through 
the  cask,  from  the  bung  to  the  most  distant  part  of  the  head. 
The  number  of  gallons  corresponding  to  the  length  of  the 
line  thus  found,  is  marked  on  the  rod.  The  logarithmic  lines 
on  the  gauging  rod  are  to  be  used  in  the  same  manner,  as 
on  the  sliding  rule. 


ULLAGE  OF  CASKS. 

129.  When  a  cask  is  partly  filled,  the  whole  capacity  is 
divided,  by  the  surface  of  the  liquor  into  two  portions;  the 
least  of  which,  whether  full  or  empty,  is  called  the  idlage. 
In  finding  the  ullage,  the  cask  is  supposed  to  be  in  one  of 
two  positions  ;  either  standing,  with  its  axis  perpendicular 
to  the  horizon  ;  or  lying,  with  its  axis  parallel  to  the  hori- 
zon. The  rules  for  ullage  which  are  exact,  particularly 
those  far  lying  casks,  are  too  complicated  for  common  use. 
The  following  are  considered  as  sufficiently  near  approx- 
imations.    See  Hution's  Mensuration. 


PROBLEM  VII. 

To  calculate  the  idlage  of  a  ST abiding  cask. 

130.  Add  together  the  squares  of  the  diameter  at  the  sur- 
face of  the  liquor,  of  the  diameter  of  the  nearest  end,  and  of 
double  the  diameter  in  the  middle  between  the  other  two; 
multiply  the  sum  by  }  of  the  distance  between  the  surface 
and  the  nearest  end,  and  the  product  by  .0028  for  ale  gallons, 
or  .0034  for  wine  gallons. 


GAUGING.  85 

If  D  =  the  diameter  of  the  surface  of  the  Hquor, 
c/=the  diameter  of  the  nearest  end, 
w  =  ihe  middle  diameter,  and 

/=■  the  distance  between  the  surface  and  the  nearest  end; 
The  ullage  in  inches=(D2  +d^  -j-Sm" )  X  |/  X  .7834. 

Ex.  If  the  diameter  at  the  surface  of  the  liquor,  in  a  stand- 
ing cask,  he  32  inches,  the  diameter  of  the  nearest  end  24, 
the  middle  diameter  29,  and  the  distance  between  the  surtace 
of  the  liquor  and  the  nearest  end  12  ;  what  is  the  ullage  f 
Ans.  27f  ale  gallons,  or  33|  wine  gallons. 


PROBLEM    VIII. 

To  calculate  the  ullage  of  a  lying  cask. 

Divide  the  distance  from  the  bung  to  the  surface  of  the 
liquor,  by  the  whole  bung  diameter,  find  the  quotient  in  the 
column  of  heights  or  versed  sines  in  a  table  of  circular  seg- 
ments, take  out  the  corresponding  segment,  and  multiply  it  by 
the  whole  capacity  of  the  cask,  and  the  product  by  li  for  the 
part  which  is  empty. 

If  the  cask  be  not  half  full,  divide  the  depth  of  the  liquor 
by  the  whole  bung  diameter,  take  out  the  segment,  multiply, 
&c.  for  the  contents  of  the  part  which  is  full. 

Ex.  If  the  whole  capacity  of  a  lying  cask  be  41  ale  gal- 
lons, or  49f  wine  gallons,  the  bung  diameter  24  inches  and 
the  distance  from  the  bung  to  the  surface  of  the  liquor  6 
inches ;  what  is  the  ullage  ? 

Ans.  7|  ale  gallons,  or  9^  wine  gallons. 


NOTES. 


Note  A.  p.    16. 

One  of  the  earliest  approximations  to  the  ratio  of  the  cir- 
wmfprfiice  of  a  circle  to  its  diameter,  was  that  o{  Archime^ 
des.  He  (iemoDstrated  tliat  the  ratio  of  the  perimeter  of  a 
regular  inscribed  prlygon  of  96  sides,  to  the  diameter  of  the 
circle,  is  greater  than  S^f  *  1  >  ^^^  that  the  ratio  of  the  pe- 
rimeter of  a  circujnscribf'd  pol'.gon  of  192  sides,  to  the  diam- 
ter,  is  less  than  3^^  :  1,  that  is,  than  22  :  7. 

JShtins  gave  the  ratio  of  355  :  115,  which  is  more  accurate 
than  any  other  expressed  in  small  numbers.  This  was  con- 
firn)ed  by  Vieta,  who  by  inscribed  and  circumscribed  poly- 
gons of  393il6  sides,  carried  the  approximation  to  ten  places 
of  figures,  viz. 

3.141592653. 

T^an  Ceulen  of  Leyden  afterwards  extended  it,  by  the  la- 
borious process  of  repeated  bisections  of  an  arc,  to  36  places. 
This  calculation  was  deemed  of  so  much  consequence  at  the 
time,  that  the  numbers  are  said  to  have  been  put  upon  his 
lomb. 

But  since  the  invention  of  ^Mjzon5,  methods  much  more 
expeditious  have  been  devised,  for  approximating  to  the  re- 
quired ratio.  These  principally  consist  in  finding  the  sum  of 
a  series,  in  which  the  length  of  an  arc  is  expressed  in  terms 
of  its  tangent. 

If  ^=the  tangent  of  an  arc,  the  radius  being  1, 

t^      t'      r      t^ 
The  arc  =^--o--|-y  — -=--{- -Q--  &c.     See  Fluxions. 


MENSURATION.  §7 

This  series  is  in  itself  very  simple.  Nothing  more  is  ne- 
cessary to  make  it  answer  the  purpose  in  practice,  than  that 
the  arc  be  smallj  so  as  to  render  the  series  sufficiently  coii- 
veraing,  and  that  the  tangent  be  expressed  in  such  siniple 
numbers,  as  can  easily  be  raised  in  the  several  powers.  The 
given  series  will  be  expressed  in  the  most  simple  numbers, 
when  the  arc  is  45°,  whose  tangent  is  equal  to  radius,  [f  the 
radius  be  1, 

The  arcof  45°  =  l-i-fi-4-hi-&c.  And  this  multi- 
plied by  8  gives  the  length  of  the  whole  circumference. 

But  a  series  in  which  the  tangent  is  smaller,  though  it  be 
less  simple  than  this,  is  to  be  preferred,  for  the  rapidity  with 
which  it  converges.  As  the  tangent  of  30''  =  v/i,  if  the  ra- 
dius be  ], 

The  arc  of  30^=^^iX{l-^^+T^-~+^-&c. 

And  this  multiplied  into  12  will  give  the  whole  circumference. 
This  was  the  series  used  by  Dr.  Halley.  By  this  also, 
Mr  Ahrnham  Sharp  of  Yorkshire  computed  the  circumfer- 
ence to  72  places  of  figurt-s,  Mr.  John  jMachin,  Professor  of 
Astronomy  in  Gresham  college,  to  lOO  places,  and  M,  De 
Lagny  to  128  place?.  Several  expedients  have  been  devised, 
by  Machin,  Euler,  Dr.  Hutton,  and  others,  to  reduce  the  la- 
bour of  summing  the  terms  of  the  series.  See  Euler's  Anal- 
ysis of  Infinites,  Hutton's  Mensuration,  Appendix  to  Maseres 
on  the  Negative  Sign,  and  Lond.  Phil.  Trans,  for  1776. 
For  a  demonstration  that  the  diameter  and  the  circumference 
of  a  circle  are  incommensurable,  see  Legendre's  Georaety, 
Note.  IV. 

The  circumference  of  a  circle  whose  diameter  is  1,  is 

3.  U 15926535,  8979323846,  2643383279, 

5028841971,  6939937510,  5820974944, 

5923078164,  0628620899,  8628034825, 

3421170679,  8214808661,  3272306647, 

09384i6  -f  or  7  - . 


8S 


NOTES. 


Note  B.  p.   17. 


The  following  multipliers  may  frequently  be  useful  ; 

l=the  side  of  an  equal  square. 

the  side  of  an  ins'bed  sq're- 

the  side  of    an    inscribed 

[equilateral  triangle. 


The  diam'rof  a  circle 


i  X.8862= 
<  X.707  =1 
{  X.866  =1 


C  X. 2821=1 

f.^  X. 2251  =  1 

(  X  .2756=1 


X.2821=the  side  of  an  equal  square. 
The  circumf.  <{  X.2251  =  thfi  side  of  an  inscribed  square. 

ithe  side  of  an  ins'bed  eq'lat.  trian. 


The  side  of  a  sq.-< 


X  l.l28=the  diam.  of  an  equ'l  circle. 
X3.545=the  circ.  of  an  equal  circle. 
X  1.414=the  dia.  of  the  circumsc.  circle. 
X  4.443= the  cir.  6f  the  circumsc.  circle. 


Note  C.  p.  19. 

The  following  approximating  rules  may  be  used  for  finding 
the  arc  of  i  circle. 

1.  The  arc  of  a  circle  is  nearly  equal  to  |  of  the  differ- 
encp  between  the  chord  of  the  whole  arc,  and  8  times  the 
chord  of  half  the  arc. 


2.  If  h^ihe  height  o(  Sin  arc,  and  c?=  the  diameter  of  the 
circle ; 

The  ^rc=2d\/,jj^  Or, 

h  3h^  S5h^ 

3.  The  zrc^2^/ dh  X  (1  -j-^^g^-f  2X5^^+ 2.4]b.7^^^-)^'' 

/     5h 
4.  The  arc=(5d|  \/  5^33^ +'*v^^^) very  nearly. 

5,  If  «=the  sine  of  an  arc,  aud  r=the  radius  of  the  circle; 

Thearc=5X(l-h  2:3;:7+  5  2.4r^  +7.2.4.6r"«  ^''• 

See  Button's  Mensuration. 


NOTES.  89 

Note  D.  p.  23. 

To  expedite  the  calculation  of  the  areas  of  circular  seg-? 
ments,  a  table  is  provided,  which  contains  the  areas  of  seg- 
ments in  a  circle  whose  diameter  is  one.  See  the  table  at  the 
end  of  the  book,  in  which  the  diameter  is  supposed  to  be  di- 
vided into  1000  equal  parts.  By  this  may  be  found  the  areas 
of  segments  of  other  circles.  For  the  heights  of  similar 
segments  of  different  circles  are  as  the  diameters.  If  then 
the  height  of  any  given  segment  be  divided  by  the  diameter 
of  the  circle,  the  quotient  will  be  the  height  of  a  similar  seg- 
ment in  a  circle  whose  diameter  is  1.  The  area  of  the  lat- 
ter is  found  in  the  table ;  and  from  the  properties  of  similar 
figures,  the  two  segments  are  to  each  other,  as  the  squares 
of  the  diameters  of  the  circles.  We  have  then  the  following 
rule : 

To  find  the  area  of  a  circular  segment  hy  the  table. 

Divide  the  height  of  the  segment  hy  the  diameter  of  the  circle  ; 
look  for  the  quotient  in  the  column  of  heights  in  the  toAle  ;  takeout 
the  corresponding  number  in  the  column  of  areas;  and  multiply  it 
by  the  square  of  the  diameter. 

It  is  to  be  observed,  that  the  figures  in  each  of  the  columns 

in  the  table  are  decimals. 

If  accuracy  is  required,  and  the  quotient  of  the  height  di- 
vided by  the  diameter,  is  between  two  numbers  in  the  column 
of  heights;  allowance  may  be  made  for  a  proportional  part 
of  the  difference  of  the  corresponding  numbers  in  the  column 
ef  areas;  in  the  same  manner,  as  in  taking  out  logarithms. 

Segments  greater  than  a  semicircle  are  not  contained  in  the 
table.  If  the  area  of  such  a  segment  is  required,  as  ABD 
(Fig.  9.),  find  the  area  of  the  segment  ABO,  and  subtract 
his   from  the  area  of  the  whole  circle. 

Divide  the  height  of  the  given  segment  by  the  diameter, 
subtract  the  quotient  from  1,  find  the  remainder  in  the  column 
of  heights,  subtract  the  corresponding  area  from  .7854  and 
multiply  this  remainder  by  the  square  of  the  diameter. 

13 


90  MENSURATION. 

Ex.   I.  What  is  the  area  of  a  segment  whose  height  is  16, 
the  diameter  of  the  circle  being  48  ?  Ans.  526. 

2.  What  is   the  area  of  a  segment  whose  height  is  32,  the 
diameter  being  48?  Ans.   1281.55. 

The  following  rules  may  also  be   used  for  a  circular  seg- 
ment. 

1.  To  the  chord  of  the  whole  arc,  add  |  of  the  chord  of 
half  the  arc,  and  multiply  the  sum  by  |  of  the  height. 

If  C  and  c=the  two  chords,  and  A=the  height ; 
The  segment  =(C+|c)fA  nearly. 

2.  U  h=  the  height  of  the  segment,  and  d={he  diameter 
of  the  circle  ; 

h        h'         h' 
Thesegment=2A>/^Ax(f-^-285^-f2^3  ^c.) 


Note  E.  p.  29. 

The  term  solidity  is  used  here  in  the  customary  sense,  to 
express  the  magnitude  of  any  geometrical  quantity  of  three 
dimensions,  length,  breadth,  and  thickness ;  whether  it  be  a 
solid  body,  or  a  fluid,  or  even  a  portion  of  empty  space. 
This  use  of  the  word,  however,  is  not  altogether  free  from 
objection.  The  same  term  is  applied  to  one  of  the  general 
properties  of  matter ;  and  also  to  that  peculiar  quality  by 
which  certain  substances  are  distinguished  (rom  fluids.  There 
seems  to  be  an  impropriety  in  speaking  of  the  solidity  of  a 
body  of  water,  or  of  a  vessel  which  is  empty.  Some  writers 
have  therefore  substituted  the  word  volume  for  solidity.  But 
the  latter  term,  if  it  be  properly  defined,  may  be  retained 
without  danger  of  leading  to  mistake. 


Note  F.  p.  35. 

The  geometrical  demonstration  of  the  rule  for  finding  the 
solidity  of  a  frustum  of  a  pyramid,  depends  on  the  following 
proposition  : 


NOTES.  91 

^ifntstum  of  a  triangular  pyramid  is  equal  to  three  pyramids  ;  the 
greatest  and  least  of  which  are  equal  in  height  to  the  frustum^  and 
have  the  fji'o  ends  of  the  frustum  for  their  bases  ;  and  the  third  is  u 
mean  proportional  betuieen  the  other  tn;o. 

Let  ABCDFG  (Fig.  34.)  be  a  frustum  of  a  triangular  py- 
ramid. If  a  plane  be  supposed  to  pass  through  the  points 
AFC,  it  will  cut  off  the  pyramid  ABCF.  The  height  of  this 
is  evidently  equal  to  the  height  of  the  frustum,  and  its  base 
is  ACB,  the  greater  end  of  the  frustum. 

Let  another  plane  pass  through  the  points  AFD.  This 
will  divide  the  remaining  part  of  the  figure  into  two  triangulai? 
pyramids  AFDG  and  AFDC.  The  height  of  the  formel^  is 
equal  to  the  height  of  the  frustum,  and  its  base  is  13FG,  the 
smaller  end  of  the  frustum.  /^ 

To  find  the  magnitude  of  the  third  pyramid  AFCD,  let  F 
be  now  considered  as  the  vertex  of  this,  and  of  the  second 
pyramid  AFDG.  Their  bases  will  then  be  the  triangles 
ADC  and  ADG.  As  these  are  in  the  same  plane,  the  two 
pyramids  have  the  same  altitude,  and  are  to  each  other  as 
their  bases.  But  these  triangular  bases,  being  between  the 
same  parallels,  are  as  the  lines  AC  and  DG.  Therefore  the 
pyramid  AFDC  is  to  the  pyramid  AFDG  as  AC  to  DG  ; 
and  AFCD'  :  AFDG' : :  AC'  :  IJG'.  (Alg.  39L)  But  the 
pyramids  ABCF  and  AFDG,  having  the  same  altitude,  are 
as  their  bases  ABC  and  DFG,  that  is,  as  AC^  and  DG^. 

(Euc.  19,  60 We  have  then     

AFDC'  :  AFDG'^jAC'^G" 
ABCF:  AFDG::  AC'  :  DG' 


Therefore  AFDC"  :  AFDG\:ABCF  :  AFDG. 

And  AFDC' =AFDGx  ABCF. 

That  is,  the  pyramid  AFDC  is  a  mean  proportional  be- 
tween AFDG  and  ABCF. 

Hence,  the  solidity  of  a  frustum  of  a  triangular  pyramid  is 
equal  to  ^  of  the  height,  multiplied  into  the  sum  of  the  areas 
of  the  two  ends  and  the  square  root  of  the  product  of  these 
areas.  This  is  true  also  of  a  frustum  of  any  other  pyramid. 
(Sup.  Euc.  12,  3.  Cor.  2.) 

If  the  smaller  end  of  a  frustum  of  a  pyramid  be  enlarged, 
till  it  is  made  equal  to  the  other  end  ;  the  frustum  will  become 
a  prism,  which  may  be  divided  into  three  equal  pyramids. 
(Sup.  Euc.  15,  3.) 


4e  MENSURATION. 


Note  G.  p.  59. 

The  following  simple  rule  for  the  solidity  of  round  timber, 
or  of  any  cylinder,  is  nearly  exact : 

Multiply  the  length  into  twice  the  square  of  ^  of  the  circumference. 

If  C=- the  circumference  of  a  cylinder; 
The  area  of  the  base=4^  =  Y2:5gg-But  2^^^^    =j^^ 

It  is  common  to  measure  hewn  timber,  by  multiplying  the 
length  into  the  square  of  the  quarter-girt.  This  gives  exact- 
ly the  solidity  of  a  parallelepiped,  if  the  ends  are  squares.  But 
if  the  ends  are  parallelograms,  the  area  of  each  is  less  than 
the  square  of  the  quarter-girt.  (Euc.  27,  6.) 

Timber  which  is  tapering  may  be  exactly  measured  by 
the  rule  for  the  frustum  of  a  pyramid  or  cone  (Art.  50,  68.) ; 
or,  if  the  ends  are  not  similar  figures,  by  the  rule  for  a  pris- 
moid.  (Art.  55,)  But  for  common  purposes,  it  will  be  suffi- 
cient to  multiply  the  length  by  the  area  of  a  section  in  the 
middle  between  the  two  ends. 


A  TABLE 


'DF  THE  SEGMENTS    OF    A    CIRCLE,    WHOSE    DIAMETER    IS    1,    AND  IS    SUP- 
POSED TO  BE  DIVIDED  INTO   1000  EQUAL  PARTS. 


Height. 

Area  Seg. 

Height. 

Area  Seg 

I  Height 

Area  Seg. 

.001 

.000042 

.034 

.008273 

.067 

.022652 

002 

000119 

035 

008638 

068 

023154 

003 

000219 

036 

009008 

069 

023659 

004 

000337 

037 

009383 

070 

024168 

005 

000471 

038 

009763 

071 

024680 

006 

000618 

039 

010148 

072 

025195 

007 

000779 

040 

010537 

073 

025714 

008 

000952 

041 

010932 

074 

026236 

009 

001135 

042 

011331 

075 

026761 

010 

001329 

043 

011734 

076 

027289 

Oil 

001533 

044 

012142 

077 

027821 

012 

001746 

045 

012554 

078 

028356 

013 

001968 

046 

012971 

079 

028894 

014 

002199 

047 

013392 

080 

029435 

015 

002438 

048 

013818 

081 

029979 

016 

002685 

049 

014247 

082 

030526 

017 

002940 

050 

014681 

083 

031076 

018 

003202 

051 

015119 

084 

031629 

019 

003472 

052 

015561 

085 

032186 

020 

003748 

053 

016007 

086 

032745 

021 

004032 

054 

016457 

087 

033307 

022 

004322 

055 

016911 

088 

033872 

023 

004618 

056 

017369 

089 

034441 

024 

004921 

057 

017831 

090 

035011 

025 

005231 

058 

018296 

091 

035585 

026 

005546 

059 

018766 

092 

036162 

027 

005867 

060 

019239 

093 

036741 

028 

006194 

061 

019716 

094 

037323 

029 

006527 

062 

020206 

095 

037909 

030 

006865 

063 

020690 

096 

038496 

031 

007209 

064 

021178 

097 

039087 

032 

007558 

065 

021659 

098 

039680 

.033 

.007913 

.066 

.022154 

.099 

.040276  1 

94 


TABLE  OF  CIRCULAR  SEGMENTS. 


1  Height. 

Area  Seg. 

Height. 

Area  Seg. 

Height. 

Area  Seg. 

.100 

.040875 

.144 

.069625 

.188 

.102334 

101 

041476 

145 

070328 

189 

103116 

102 

042080 

146 

071033 

190 

103900 

103 

042687 

147 

071741 

191 

104685 

104 

043296 

148 

072450 

192 

105472 

105 

043908 

149 

073161 

193 

106261 

106 

044522 

150 

073874 

194 

107051 

107 

045139 

151 

074589 

195 

107842 

108 

045759 

152 

075306 

196 

108636 

109 

046381 

153 

076026 

197 

109430 

110 

047005 

154 

076747 

198 

110226 

111 

047632 

155 

077469 

199 

111024 

112 

048262 

156 

078194 

200 

111823 

113 

048894 

157 

078921 

201 

112624 

114 

049528 

158 

079649 

202 

113426 

115 

050165 

159 

080380 

203 

114230 

116 

050804 

160 

081112 

204 

115035 

117 

051446 

161 

081846 

205 

115842 

118 

052090 

162 

082582 

206 

116650 

119 

052736 

163 

083320 

207 

117460 

120 

053385 

164 

084059 

208 

118271 

121 

054036 

165 

084801 

209 

119083 

122 

054689 

166 

085544 

210 

119897 

123 

055345 

167 

086289 

211 

120712 

124 

056003 

168 

087036 

212 

121529 

125 

056663 

169 

087785 

213 

122347 

126 

057326 

170 

088535 

214 

123167 

127 

057991 

171 

089287 

215 

123988 

128 

058658 

172 

090041 

216 

124810 

129 

059327 

173 

090797 

217 

125634 

130 

059999 

174 

091554 

218 

126459 

131 

060672 

175 

092313 

219 

127285 

132 

061348 

176 

093074 

220 

128113 

133 

062026 

177 

093836 

221 

128942 

134 

062707 

178 

094601 

222 

129773 

135 

063389 

179 

095366 

223 

130605 

136 

064074 

180 

096134 

224 

131438 

137 

064760 

181 

096903 

225 

132272 

138 

065449 

182 

097674 

226 

133108 

139 

066140 

183 

098447 

227 

133945 

140 

066833 

184 

099221 

228 

,  134784 

141 

067528 

185 

099997 

229 

135624 

142 

068225 

186 

100774 

230 

136465 

.143 

.068924 

;}.''  . 

.101553  1 

.231 

,.137307  1 

TABLE  OF  CIRCULAR  SEGMENTS. 


95 


1  Height 

Area  Seg. 

Height 

Area  Seg 

Height 

Aaea.  Seg. 

232 

.138150 

.277 

.177330 

.322 

.218533 

233 

138995 

278 

178225 

323 

219468 

234 

139841 

279 

179122 

324 

220404 

235 

140688 

280 

180019 

325 

221340 

236 

141537 

281 

180918 

326 

222277 

237 

142387 

282 

181817 

327 

223215 

238 

143238 

283 

182718 

328 

224154 

239 

144091 

284 

183619 

329 

225093 

240 

144944 

285 

184521 

330 

226033 

241 

145799 

286 

185425 

331 

226974 

242 

146655 

287 

186329 

332 

227915 

243 

147512 

288 

187234 

333 

228858 

244 

143371 

289 

188140 

334 

229801 

245 

1492.30 

290 

189047 

335 

230745 

246 

150091 

291 

189955 

336 

231689 

247 

150953 

292 

190864 

337 

232634 

248 

151816 

293 

191775 

338 

233580 

249 

152680 

294 

192684 

.  339 

234526 

250 

153546 

295 

193596 

340 

235473 

251 

154412 

296 

194509 

341 

236421 

252 

155280 

297 

195422 

342 

237369 

253 

156149 

298 

196337 

343 

238318 

254 

157019 

299 

197252 

314 

239268 

255 

157890 

300 

198168 

345 

240218 

256 

158762 

301 

199085 

346 

241169 

257 

159636 

302 

200003 

347 

242121 

258 

160510 

303 

200922 

348 

243074 

259 

161386 

304 

201841 

349 

244026 

260 

162263 

305 

202761 

350 

244980 

261 

163140 

306 

203683 

351 

245934 

262 

164019 

307 

204605 

352 

246889 

263 

164899 

308 

205527 

353 

247845 

264 

165780 

309 

206451 

354 

248801 

265 

166663 

310 

207376 

355 

249757 

266 

167546 

311 

208301 

356 

250715 

267 

168430 

312 

209227 

357 

251673 

268 

169315 

313 

210154 

358 

252631 

269 

170202 

314 

211082 

359 

253590 

270 

171089 

315 

212011 

360 

254550 

271 

171978 

316 

212940 

361 

255510 

272 

172867 

317 

213871 

362 

256471 

273 

173758 

318 

214802 

363 

257433 

274 

174649 

319 

215733 

364 

258395 

275 

175542 

320 

216666 

365 

259357 

.276 

.176435 

.321 

.217599 

.366 

,  .260320 

96 


TABLE  OF  CIRCULAR  SEGMENTS. 


Height 

^  rea  Seg. 

Heigiit 

Area  &.„ 

Height. 

Area  5r"eg. 

.367 

.261284 

.412 

.305155 

.457 

.349752 

368 

262248 

413 

306140 

458 

350748 

369 

263213 

414' 

307125 

459 

351745 

370 

264178 

415 

308110 

460 

352742 

371 

265144 

416 

309095 

461 

353739 

372 

266111 

417 

310081 

462 

354736 

373 

267078 

418 

311068 

463 

355732-' 

374 

268045 

419 

312054 

464 

356730 

375 

269013 

420 

313041 

465 

357727 

376 

269982 

421 

314029 

466 

358725 

377 

270951 

422 

315016 

467 

359723 

378 

271920 

423 

316004 

468 

360721 

379 

272890 

424 

316992 

469 

361719 

380 

273861 

425 

317981 

470 

362717 

381 

274832 

426 

318970 

471 

363715 

282 

275803 

427 

319959 

472 

364713 

383 

276775 

428 

320948 

473 

365712 

384 

277748 

429 

321938 

474 

.366710 

385 

278721 

430 

322928 

475 

367709 

386 

279694 

431 

323918 

476 

368708 

387 

280668 

432 

324909 

477 

369707 

388 

281642 

433 

325900 

478 

370706 

389 

282617 

434 

326892 

479 

371705 

390 

283592 

435 

327882 

480 

372704 

391 

284568 

436 

328874 

481 

373703 

392 

285544 

437 

329866 

482 

374702 

393 

286521 

438 

330858 

483 

375702 

394 

287498 

439 

331850 

484 

376702 

395 

288476 

440 

332843 

485 

377701 

396 

289454 

441 

333836 

486 

378701 

397 

290432 

442 

334829 

487 

379700 

398 

291411 

443 

335822 

488 

380700 

399 

292390 

444 

336816 

489 

381699 

400 

293369 

445 

337810 

490 

382699 

401 

294349 

446 

338804 

491 

383699 

402 

295330 

447 

339798 

492 

384699 

403 

296311 

448 

340793 

493 

385699 

404 

297292 

449 

341787 

494 

386699 

405 

298273 

450 

342782 

495 

387699 

406 

299255 

451 

343777 

496 

388699 

407 

300238 

452 

344772 

497 

389699 

408 

301220 

453 

345768 

498 

390699 

409 

302203 

454 

346764 

499 

391699 

410 

303187 

455 

347759 

.500 

.392699 

.411 

.304171 

.456 

.348755 

_ 

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rLU. 


-M  Jiraf&n  JV. 


»IEWST'RAT10jr. 


THE 


MATHExMATICAL  PRINCIPLES 


NAVIGATION  AND  SURVEYING, 

WITH    THE 

MENSURATION 

OF  ; 

HEIGHTS  AND  DISTANCES. 

BEING 

THB  rOU&TB  PART 

OF 

.4  COURSE  OP  MATHEMATICS. 

ADAPTED    TO    THE    METHOD    OF    INSTRUCTION    IN 
THE    AMERICAN    COLLEGES. 


BY  JEREMIAH  U  VY.  1).  /.  LL.D. 

President  of  Yale  College. 


THE  SECOND  EDITION. 
NEWHAVE^T: 

miNTED    AND    PUBLISHED    BY   S.    CONVERSE. 


1824. 


DISTRICT  OF  CO^rjVECTICUT,  ss. 
Be  it  remembered  ;  That  on  the  twentieth  day  olWIarch, 
in  the  forty-first  year  of  the  Independence  ol  (he  United 
States  of  America,  Jeremiah  Day,  of  the  said  District,  h^ith 
deposited  in  this  Office,  the  title  of  a  book,  the  right  whereof 
he  claims  as  Author,  in  the  words  following,  to  wit  : 
"The  Mathematical  Principles  of  Navigation  and  Surveying,  with  the 
Mensuration  of  Heights  and  Distances.  Being  the  fourth  part  of  a  Course  of 
Mathematics,  adapted  to  the  method  of  instruction  in  the  American  Colleges. 
By  Jeremiah  Day,  Professor  of  Mathematics  and  Natural  Philosophy,  in  Yale 
College." 

In  conformity  to  the  Act  of  the  Congress  of  the  United  State?,  entitled  "  An 
Act  for  the  encouragement  of  learning,  by  securing  the  copies  of  Maps,  Charts 
and  Books,  to  the  authors  and  proprietors  of  such  copies,  during  the  times 
therein  mentioned." 

HENRY  W.  EDWARDS,  Clerk  of  the 

District  of  Connecticut. 
A  true  copy  of  Record,  examined  and  sealed  by  me, 

H.  W.  EDWARDS,  Clerk  of  the 

District  of  Connecticut. 


As  the  following  treatise  has  been  prepared  for  the  use  of  a 
class  in  College,  it  does  not  contain  all  the  details  which 
would  be  requisite  for  a  practical  navigator  or  surveyor. 
The  object  of  a  scientific  education  is  rather  to  teach  princi- 
pies,  than  the  minute  rules  which  are  called  for  in  profession- 
al practice.'  The  principles  should  indeed  be  accompanied 
with  such  illustrations  and  examples  as  will  render  it  easy 
for  the  student  to  make  the  applications  for  himself  when- 
ever occasion  shall  require.  But  a  collection  of  rules 
merely,  would  be  learned  only  to  be  forgotten,  except  by  a 
few  who  might  have  use  for  them  in  the  course  of  their  bu- 
siness. There  are  many  things  belonging  to  the  art  of  navi- 
gation, which  are  not  comprehended  in  the  mathematical 
part  of  the  subject.  Seamen  will  of  course  make  use  of  the 
valuable  system  of  Mackay,  or  the  still  more  complete  work 
ofBowditch. 

The  student  is  supposed  to  be  familiar  with  the  principles 
of  Geometry  and  Trigonometry,  before  he  enters  upon  the 
present  number,  which  contains  little  more  than  the  applica- 
tion of  those  principles  to  some  of  the  most  simple  problems 
in  heights  and  distances,  navigation,  and  surveving. 


CONTENTS. 


Page. 

Heiohts  and  Distances      -     -     .     _     .  j 

NAVIGATION. 

Section  I.     Plane  Sailing        ---..--_  15 

11.     Parallel  and  Middle  Latitude  Sailing      -  25 

III.  Mercator's  Sailing 32 

IV.  Traverse  Sailing  -----.-.  40 
V.     Miscellaneous  Articles 

The  Plane  Chart -  45 

Mercator's  Chart 46 

Oblique  Sailing -     -  47 

Current  Sailing      -------  49 

Hadle)''s  Quadrant      ------  51 


SURVEYING. 

Section  f.     Surveying  a  field  by  measuring  round  it  -  57 

II      Oilier  methods  of  surveying     -     -     -     -  73 

III.  Dividing  and  laying  out  lands  -     -     -     -  62 

IV.  Levelling 86 

V.     The  Magnetic  Needle 92 

Notes 95 

Tables 109 


HEIGHTS  AND  DISTANCES. 


Art.  1.  rWlHE  most  direct  and  obvious  method  of  determin- 
A  ing  the  distance  or  height  of  any  object,  is  to  ap- 
ply to  it  some  known  measure  of  length,  as  a  foot,  a  yard,  or 
a  rod.  In  this  manner,  the  height  of  a  room  is  found,  by  a 
joiner's  rule ;  or  the  side  of  a  tield  by  a  surveyors  chain. 
Eut  in  many  instances,  the  object,  or  a  part,  at  least  of  the 
line  which  is  to  be  measured  is  inaccessible.  We  may  wish 
to  determine  the  breadth  of  a  river,  the  height  of  a  cloud,  or 
the  distances  of  the  heavenly  bodies.  In  such  cases  it  is  ne- 
cessary to  measure  some  other  line  ;  from  which  the  requir- 
ed line  may  be  obtained,  by  geometrical  construction,  or 
more  exactly,  by  trigonometrical  calculation.  The  line  first 
measured  is  frequently  called  a  base  line. 

2.  In  measuring  angles,  some  instrument  is  used  which 
contains  a  portion  of  a  graduated  circle  divided  into  degrees 
and  minutes.  For  the  proper  measure  of  an  angle  is  an  arc  of 
a  circle,  whose  centre  is  the  angular  point.  (Trig.  74.)  The 
instruments  used  for  this  purpose  arc  made  in  different 
forms,  and  with  various  appendages.  The  essential  parts 
are  a  graduated  circle,  and  an  index  with  sight-holee,  for  tak- 
ing the  directions  of  the  lines  which  include  the  angles. 

-  3.  Angles  of  elevation,  and  of  depression,  are  in  a  plane 
perpf^ndicular  to  the  horizon,  which  is  called  a  vertical  plane. 
An  anele  of  elevation  is  contained  between  a  parallel  to  the 
horizon,  and  an  ascending  line,  as  BAG.  (Fig.  2.)  An  angle 
o(  depression  is  contained  between  a  parallel  to  the  horizon, 
and  a  descending  line,  as  DCA.  The  complement  of  this  is 
the  angle  ACB. 

4.  The  instrument  by  which  angles  of  elevation,  and  of 
depression,  are  commonly  measured,  is  called  a  Quadrant. 
In  its  most  simple  form,  it  is  a  portion  of  a  circular  board 


2  MENSURATION  OF 

ABC,  (Fig.  1.)  on  which  is  a  graduated  arc  of  90  degree?, 
AB,  a  plumb  line  CP,  suspended  from  the  central  point  C, 
and  two  sight-holes  D  and  E,  for  taking  the  direction  of  the 
object. 

To  measure  an  angle  o( elevation  with  this,  hold  the  plane 
of  the  instrument  perpendicular  to  the  horizon,  bring  the  cen- 
tre C  to  the  angular  point,  and  direct  the  edge  AC  in  such  a 
mattner,  that  the  object  G  may  be  seen  through  the  two 
sight-holes.  Then  the  arc  BO  measures  the  angle  BCO, 
which  is  equal  to  the  angle  of  elevation  FCG.  For  as  the 
plumb-line  is  perpendicular  to  the  horizon,  the  angle  FCO 
is  a  right  angle,  and  therefore  equal  to  BCG.  Taking  from 
these  the  common  angle  BCF,  there  will  remain  the  angle 
BCO=FCG. 

In  taking  an  angle  o^  depression,  as  HCIj  (Fig.  1.)  the  eye 
is  placed  at  C,  so  as  to  view  the  object  at  L,  through  the 
sight-holes  D  and  E. 

5.  In  treating  of  the  mensuration  of  heights  and  distances, 
no  new  principles  are  to  be  brought  into  view.  We  have 
only  to  mak^  an  application  of  the  rules  for  the  solution  of 
triangles,  to  the  particular  circumstances  in  which  the  obser- 
ver may  be  placed,  with  respect  to  the  line  to  be  measured. 
These  are  so  numerous,  that  the  subject  may  be  divided  inta 
a  great  number  of  distinct  cases.  But  as  they  are  all  solved 
upon  the  same  general  principles,  it  will  not  be  necessary  to 
give  examples  under  each.  The  following  problems  may 
serve  as  a  specimen  of  those  which  most  frequently  occur  in 
practice 

Problem  I. 

To  FIND  THE  PERPENDICULAR  HEIGHT  OF  AN  ACCESSIBLE  OB- 
JECT STANDING  ON  A  HORIZONTAL  PLANE. 

6.  MEASURE  FROM  THE  OBJECT  TO  A  CONVENIENT  STATION, 
AND  THERE  TAKE  THE  ANGLE  OF  ELEVATION  SUBTENDED  BY 
THE  OBJECT. 

If  the  distance  AB  (Fig.  2.)  be  measured,  and  the  angle 
of  elevation  BAC;  there  will  be  given  in  the  right  angled 
triangle  ABC,  the  base  and  the  angles,  to  find  the  perpen- 
dicular. (Trig.  137.) 

As  the  instrument  by  which  the  angle  at  A  is  measured,  is 
commonly  raised  a  few  feet  above  the  ground ;  a  point  B 
must  be  taken  in  the  object,  so  that  AB  shall  be  parallel  to 


HEIGHTS  AND  DISTANCES.  3 

the  horizon.     The  part  BP,  may  afterwards  be  added  to  the 
heia:ht  EC,  found  by  trigonometrical  calculation. 

Ex.  1.  What  is  the  height  of  a  tower  BC,  (Fig.  2.)  if  the 
distance  AB,  on  a  horizontal  plane,  be  9S  feet;  and  the  an- 
gle BAC  35|  degrees  ? 

Makinor  the  hypothenuse  radius,  (Trig.  121.) 
Cos  BAC  :  AB:  :Sin  BAC  :  BC=69.9  feet. 

For  the  geometrical  constructio7i  of  the  problem,  see  Trig. 
169. 

2.  What  is  the  height  of  the  perpendicular  sheet  of  water 
at  the  falls  ofNiagara.  if  it  subtends  an  angle  of  40  degrees, 
at  the  distance  of  163  feet  from  the  bottom,  measured  on  a 
horizontal  plane?  Ans.  136 J  feet. 

7.  If  the  height  of  the  object  be  known,  its  distance  may  be 
found  by  the  angle  of  elevation.  In  this  case  the  angles,  and 
the  perpendicular  of  the  triangle  are  given,  to  find  the  base. 

Ex.  A  person  on  shore,  taking  an  observation  of  a  ship's 
mast  which  is  known  to  be  99  feet  high,  finds  the  angle  of 
elevation  3^  degrees.  What  is  the  distance  of  the  ship  from 
the  observer  ?  Ans.  98  rods. 

8.  If  the  observer  be  stationed  at  the  top  of  the  perpen- 
dicular BC,  (Fig.  2.)  whose  height  is  known  ;  he  may  find 
the  length  of  the  base  line  AB,  by  measuring  the  angle  of 
depression  ACD,  which  is  equal  to  BAC 

Ex.  A  seaman  at  the  top  of  a  mast  G6  feet  high,  looking  at 
another  ship,  finds  the  angle  of  depression  1 0  degrees.  What 
is  the  distance  of  the  two  vessels  from  each  other  ? 

Ans.  22f  rods. 

We  may  find  the  distance  between  tzoo  objects  which  are 
in  the  same  vertical  plane  with  the  perpendicular,  by  calcu- 
lating the  distance  of  each  from  the  perpendicular.  Thus 
AG  (Fig.  2.)  is  equal  to  the  difference  between  AB  and  GB. 

Problem.  H. 

to  find  the  height  of  an  accessible  object  standing 
dn  an  inclined  plane. 

9.  Measure  the  distance  from  the  object  to  a  con- 
venient STATION,  AND  TAKE  THE  ANGLES  WHICH  THIS  BASE 
MAKES  WITH  LINES  DRAWN  FROM  ITS  TWO  ENDS  TO  THE  TOP 
OF  THE  OBJECT. 


4  MENSURATION  OF 

If  the  base  AB  ''Fig.  3.)  be  measured,  and  the  angles  BAG 
and  ABC  ;  there  will  be  given,  in  the  oblique  angled  triangle 
ABC,  the  side  AB  and  the  angles,  to  find  BC.     (Trig.  150.) 

Or  the  height  BC  maj  be  found  by  measuring  the  distan- 
ces BA,  AD,  and  taking  the  angles  BAC  and  BDC.  There 
will  then  be  given  in  the  triangle  ADC,  the  angles  and  the 
side  AD,  to  find  AC;  and  consequently,  in  the  triangle 
ABC,  the  sides  AB  and  AC  with  the  angle  BAC,  to  find 
BC. 

Ex.  If  AB  (Fig.  3.)  be  70  (eet,  the  angle  B  101°  25' 
and  the  angle  A  44®  42' ;  what  is  the  heiglit  of  the  tree 
BC? 

SinC  :  AB::SinA  :  BC=95.0feet. 

For  the  geometrical  construction  of  the  problem,  see  Trigc 
169. 

10.  The  following  are  some  of  the  methods  by  which  the 
height  of  an  object  may  be  found,  without  measuring  the  an- 
gle of  elevation. 

1.  Bi^  shadows.  Let  the  staff  be  (Fig.  4.)  be  parallel  to 
an  object  BC  whose  height  is  required.  If  the  shadow  of 
BC  extend  to  A,  and  that  of  6c  to  a  ;  the  rays  of  light  CA 
and  ca  coming  from  the  sun  may  be  considered  parallel  ;  and 
therefore  the  triangles  ABC  and  aba  are  similar ;  so  that 

ab  :  bc'.AB  :  BC. 

Ex.  \(ah  be  3  feet,  be  5  feet,  and  AB  G9  Cect,  what  is  the 
height  of  BC  ?  Ans.  1 1 5  (eet. 

2.  By  parallel  rods.  If  two  poles  am  and  en  (Fig.  5.)  be 
placed  parallel  to  the  object  BC,  and  at  such  distances  as  to 
bring  the  points  C,  c,  0  in  a  line,  and  \(  ab  be  made  parallel 
to  AB;  the  triangles  ABC,  and  abe  will  be  similar  ;  and  we 
shall  have 

ab  :k::AB:  BC. 

One  pole  will  be  sufficient,  if  the  observer  can  place  his  eye 
at  the  point  A,  so  as  to  bring  A,  a,  and  C  in  a  line. 

3.  By  a  mirror.  Let  the  smooth  surface  of  a  body  of 
water  at  A,  (Fig.  G.)  or  any  plane  mirror  parallel  to  the  hori- 
zon, be  so  situated,  that  the  eye  of  the  observer  at  c  may 
view  the  top  of  the  object  C  reflected  from  the  mirror.  By 
a  law  of  Optics,  the  angle  BAC  is  equal  to  bAe ;  aod  if  be 


HEIGHTS  AND  DISTANCES.  f. 

be  made  parallel  to  BC,  the  triangle  bAc  will  be  similar  to 
BAG  :  ?o  that 

A6 :  bc::AB  :  BC. 


Problem  HI. 

to  ri.vd  the  height  of  an  inaccessible  object  above  a 
horizontal  plane. 

11.  Take  two  stations  in  a  vertical  plane  passing 
through  the  top  of  the  object,  measure  the  distance 
from  one  station  to  the  other,  and  the  angle  of  eleva- 
tion at  each. 

If  the  base  AR  (Fig.  7.)  be  measured  with  the  angles  CBP 
and  CAB ;  as  ARC  is  the  supplement  of  CRP,  there  will  be 
given,  in  the  oblique  angled  triangle  ARC,  the  side  AB  and 
the  angles,  to  fmd  BC;  and  then,  in  the  right  angled  triangle 
BCP,  the  hypothenuse  and  the  angles,  to  find  the  perpendic- 
ular CP. 

Ex.  I.  If  C  (Fig.  7.)  be  the  top  of  a  spire,  the  horizontal 
base  line  AB  100*  feet,  the  angle  of  elevation  BAC  40°, 
and  the  angle  PBC  60°  ;  what  is  the  perpendicular  height  oj; 
the  spire  ? 

The  difference  between  the  angles  PBC  and  BAC  is  equal 
to  ACB.     (Euc.  32.  1) 

Then  Sin  ACB  :  AR::Sin  BAC  :  BC  =  187.9 
And  R  :  BC::Sin  PBC  :  CP=162|  feeU 

2.  If  two  persons  120  rods  from  each  other,  are  standing 
on  a  horizontal  plane,  and  also  in  a  vertical  plane  passing 
through  a  cloud,  both  being  on  the  same  side  of  the  cloud  : 
and  if  they  find  the  angles  of  elevation  at  the  two  stations  to 
be  G8=  and  76°  ;  what  is  the  height  of  the  cloud  ? 

Ans.  2  miles  135.7  rods. 

12.  The  preceding  problems  are  useful  in  particular  ca- 
ses. But  the  following  is  2i  general  rule,  which  may  be  used 
for  finding  the  height  of  any  object  whatever,  within  mode- 
rate distances. 


6  BIENSURATION  OF 

Problem  IV. 

to  find  tht:  height  of  any  object^  bv  observations  at 
two  stations. 

13.  Measure  the  base  line  between  the  two  stations, 
the  angles  between  this  ease  and  lines  drawn  from  each 
of  the  stations  to  each  end  of  the  object,  and  the  an- 
gle subtended  by  the  object,  at  one  of  the  stations. 

IfBC(Fig.  8.)  be  the  object  whose  height  is  required, 
and  if  the  distance  between  the  stations  A  and  D  be  meas- 
ured, with  the  angles  ADC,  DAC,  ADB,  DAB,  and  BAG; 
there  will  be  given,  in  the  triangle  ADC,  the  side  AD  and 
the  angles,  to  find  AC  ;  in  the  triangle  ADB,  the  side  AD 
and  the  angles,  to  find  AB;  and  then,  in  the  triangle  BAG, 
the  sides  AB  and  AC  with  the  included  angle,  to  find  the  re- 
quired height  BC. 

If  the  two  stations  A  and  D  be  in  the  same  plane  with  BC, 
the  angle  BAC  will  be  equal  to  the  ditTerence  between  BAD 
and  CAD.  In  this  case  it  will  not  be  necessary  to  measure 
BAC. 

Ex.     If  AD=83  feet,  (Fig.  8.)  (  ADB=33°, 
(  ADC=51°  (DAB=121° 

iDAC  =  95°  BAC=26^ 

What  is  the  height  of  the  object  BC  ? 
SinACD  :  AD:: ADC  :  AC  =  n5.3 
Sin  ABD  :  AD: :  ADB  :  AB=103.1 
(AC+AB)  :  (AC-AB)::Tan  i(ABC-l-ACB)  :  Tani(ABC 
-ACB)  =  13°38' 
Sin  ACB  :  AB:  :Sin  BAC  t  BC=50.57  i^Qi. 
If  the  object  BC  be  perpendicular  to  the  horizon,   its 
height,  after  obtaining  AB  and  BC  as  before,  may  be  found 
by  taking  the  angles  o(  elevation  BAP  and  CAP.    The  differ- 
ence of  the  perpendiculars  in  the  right  angled  triangles  ABP 
and  ACP,  will  be  the  height  required. 

Problem  V. 

to  find  the  distance  of  an  inaccessible  object. 

14.  Measure  a  base  line  between  two  stations,  and 
the  angles  between  this  and  lines  drawn  from  each 
of  the  stations  to  the  object. 


HEIGHTS  AND  DISTANCES.  7 

If  C  (Fig,  9.)  be  the  object,  and  if  the  distance  between 
the  stations  A  and  B  be  measured,  with  the  angles  at  B  and 
A;  there  will  be  given,  in  the  oblique  angled  triangle  ABC, 
the  side  AB  and  the  angles,  to  find  AC  and  BC,  the  distan- 
ces of  the  object  from  the  two  stations. 

For  the  geometrical  construction,  see  Trig.  169. 
Ex.  1.     What  are  the  distances  of  the  two  stations  A  and 
B  (Fig.  9.)  from  the  house  C,  on  the  opposite  side  of  a  river; 
if  AB  be  26.G  rods,  B  92^  46',  and  A  ^8"  40' .? 

The  angle  C  =  180-(A-hB)=48^  34'.     Then 
Q-     r-  .  ATJ..  ^Sin  A  :BC=22.17 
Sm  CAB,.  (sinB:AC=35.44 

2.  Two  ships  in  a  harbour,  wishins:  to  ascertain  how  far 
they  are  from  a  fort  on  shore,  find  that  their  mutual  distance 
is  90  rods,  and  that  the  angles  formed  between  a  line  from 
one  to  the  other,  and  lines  drawn  from  each  to  the  fort  are 
4^°  and  56°  15'.  What  are  their  respective  distances  from 
the  fort.'*  Ans.  76.3  and  64.9  rods. 

15.  The  perpendicular  distance  of  the  object  from  the 
line  joining  the  two  stations  may  be  easily  found,  after  the  dis- 
tance from  one  of  the  stations  is  obtained.  The  perpendicu- 
lar distance  PC  (Fig  9  J  is  one  of  the  sides  of  the  right  an> 
gled  triangle  BCP.     Therefore 

R:BC::SinB  :  PC 

Problem  VI. 

to  find  the  distance  between  two  objects,  when 
the  passage  fuom  one  to  the  other,  in  a  straight 
line,  is  obstructf-d. 

16.  Measure  the  right  lines  from  one  station  to 
each  of  the  objects,  and  the  angle  included  between 
these  lines. 

If  A  and  B  (Fig.  10.)  be  the  two  objects,  and  if  the  dis- 
tances BC  and  AC  be  measured,  with  the  angle  at  C;  there 
will  be  given,  in  the  oblique  angled  triangle  ABC.  two  sides 
and  the  included  angle  to  find  the  other  two  angles,  and  the 
remaining  side.     (Trig.  153.) 

Ex.  The  pai^sage  between  the  two  objec  tsA  and  B  (Fig 
10.)  being  obstructed  by  a  morass,  the  line  BC  was  measured 
and  found  to  be  109  rods,  the  line  AC  76  rods,  and  the  angle 
at  C  lOl'^  30'.     What  is  the  di^:tance  AB  ^ 

Ans.   144.7  rods. 


MENSURATION  OF 


Problem  VII. 

to  find  the  distance  between  two  inaccessible  objects. 

17.  Measure  a  base  line  between  two  stations  and 

THE  angles  between  THIS  BASE  AND  LINES  DRAWN  FROM 
EACH  OF  THE  STATIONS,  TO  EACH  OF  THE  OBJECTS. 

If  A  and  BfFig.  11.)  be  the  two  objects,  and  if  the  distance 
between  the  stations  C  and  D  be  measured,  with  the  angles 
BDC,  BCD,  ADC,  and  ACD  ;  the  lines  AC  and  BC  may 
be  found  as  in  Problem  V,  and  then  the  distance  AB  as  in 
Problem  VI. 

This  rule  is  substantially  the  same  as  that  in  art.  13.  The 
two  stations  are  supposed  to  be  in  the  same  plane  with  the 
objects.  If  they  are  not,  it  will  be  necessary  to  measure  the 
angle  ACB. 

18.  The  same  process  by  which  we  obtain  the  distance  of 
two  objects  from  each  other,  will  enable  us  to  find  the  dis- 
tance between  one  of  these  and  a  third,  between  that  and  a 
fourth,  and  so  on,  till  a  connection  is  formed  between  a  great 
number  of  remote  points.  This  is  the  plan  of  the  great  Trig- 
onometrical  Surveys  which  have  been  lately  carried  on,  with 
surprising  exactness,  particularly  in  England  and  France. 
See  Surveying,  Section  II. 

19.  In  the  preceding  problems  for  determining  altitudes, 
the  objects  are  supposed  to  be  at  such  moderate  distances, 
that  the  observations  are  not  sensibly  affected  by  the  spherical 
figure  of  the  earth.  The  height  of  an  object  is  measured 
from  an  horizontal  plane,  passing  through  the  station  at  which 
the  angle  of  elevation  is  taken.  But  in  an  extent  of  several 
miles,  the  figure  of  the  earth  ought  to  be  taken  into  account. 

Let  AB  (Fig.  12)  be  a  portion  of  the  earth's  surface,  H 
an  object  above  it,  and  AT  a  tangent  at  the  point  A,  or  a 
horizontal  line  passing  through  A.  Then  HT,  the  oblique 
height  of  the  object  above  the  horizon  of  A,  is  only  a  part  of 
the  height  above  the  surface  of  the  earth,  or  the  level  of  the 
ocean.  To  obtain  the  true  altitude,  it  is  necessary  to  add 
BT  to  the  height  HT  found  by  observation.  The  height 
BT  may  be  calculated,  if  tiie  diameter  of  the  earth  and  the 
distance  AT  be  previously  known.  Or  if  the  height  BT  be 
first  determined  from  observation,  with  the  distance  AT ;  the 
diameter  of  the  earth  mav  be  thence  deduced. 


HEIGHTS  AND  DISTANCES, 


PUOBLEM  VIH. 

to  find  the  diameter  of  the  earth,  from  the  known 
height  of  a  distant  mountain,  whose  summit  is  just  vis- 
ible in  the  horizon. 

20.  From  the  square  of  the  distance  divided  by  the 
height,  subtract  the  height. 

IfBT   (Fig.  12)  be  a   mountain    whosn   height  is   known, 
with  the  distance  AT;  and  if  the  sanimit  T  be  just  visible  in 
the  horizon  at  A  ;  then  AT  is  a  tangent  at  the  point  A. 
Let  2BC  =  D,  the  diameter  of  the  earth, 
AT  =  c^,  (he  distance  of  the  mountain, 
BT=/i,  its  height. 

Then  considering  AT  as  a  straight  line,  and  the  earth  as  a 
sphere,  we  have  (Euc.  36.  3.) 

(2BC  +  BT)xBT  =  AT'  ;  Ihat  is,  (D+/0  X/^=£/^ 
and  reducing  the  equation, 

D=^-A 
A 

Ex»  The  highest  point  of  the  Andes  is  about  4  miles  above 

ihe  level  of  the  ocean.     If  a  straight  line  from  this  touch  the 

surface  of  the  water  at  the    distance  of  178}  miles;  what  is 

the  diameter  of  the  earth  ?  Ans.  7940  miles. 

21.  If  the  distance  AT  (Fig.  12.)  be  unknowny  it  may  be 
found  by  measuring  with  a  quadrant  the  angle  ATC.  Draw 
BG  perpendicular  to  BC  ;  and  join  CG.  The  triangles  ACG 
and  BCG  are  equal,  because  each  has  a  right  angle,  the  sides 
AC  and  BC  are  equal,  and  the  hypothenuse  CG  is  common. 
Therefore  BG  and  AG  are  equal.  In  the  right  angled  trian- 
gle BGT,  the  angle  BTG  is  given,  and  the  perpendicular 
BT.  From  these  may  be  found  BG  and  TG,  whose  sum  is 
equal  to  AT,  the  distance  required.* 

22.  In  the  common  measurement  of  angles,  the  light  is 
supposed  to  come  from  the  object  to  the  eye  in  a  straight 
line.  But  this  is  not  strictly  true.  The  direction  of  ihc  light 
is  atfected  by  the  refraction  of  the  atmosphere.  If  the  object 
be  near,  the  deviation  is  very  inconsiderable.     But  in  an  cx- 

•  This  method  of  determining  the  diameter  of  the  earth  is  not  as  accurate 
aa  that  by  measuring;  a  de^^ree  of  Latitude,    See  Surveying,  Sec.  I  J. 


10  MENSURATION  OF 

tent  of  several  miles,  and  particularly  in  such  nice  observa- 
tions as  determining  the  height  of  distant  mountains,  and  the 
diameter  of  the  earth,  it  is  necessary  to  make  allowance  for 
the  refraction.* 

Problem  IX. 

To  FIND  THE  GREATEST  DISTANCE  AT  WHICH  A  GIVEN  OB- 
JECT CAN  BE  SEEN  ON  THE  SURFACE  OF  THE  EARTH. 

23.  To  THE  PRODUCT  OF  THE  HEIGHT  OF  THE  OBJECT  INTO 
THE  DIAMETER  OF  THE  EARTH,  ADD  THE  SQ,UARE  OF  THE 
HEIGHT  ;  AND  EXTRACT  THE  SQUARE  ROOT  OF  THE  SUM. 

Let  2BC  =  D,  the  diameter  of  the  earth,  (Fig.  12.) 
BT=/f,  the  height  of  the  object, 
AT  =  f/,  the  distance  required. 
Then  (D+^)XA=</-.     And  d=^/Dh-^hK 
Ex.  If  the  diameter  of  the  earth  be  7940  miles,  and  Mount 
JEtna  2  miles  high  ;  how  far  can  its  summit  be  seen  at  sea  ? 

Ans.  126  miles. 
The  actual  distance  at  which  an  object  can  be  seen,  is  in- 
creased by  the  refraction  of  the  air.* 

24.  In  this  problem,  the  eye  is  supposed  to  be  placed  at 
the  level  of  the  ocean.  But  if  the  observer  be  elevated  above 
the  surface,  as  on  the  deck  of  a  ship,  he  can  see  to  a  greater 
distance.  If  BT  (Fig.  13.)  be  the  height  of  the  object,  and 
B  'T'  the  height  of  the  eye  above  the  level  of  the  ocean  ;  the 
distance  at  which  the  object  can  be  seen,  is  evidently  equal 
to  the  sum  of  the  tangents  AT  and  AT'. 

Ex.  The  top  of  a  ship's  mast  1 32  feet  high  is  just  visible  in 
the  horizon,  to  an  observer  whose  eye  is  S3  feet  above  the 
surface  of  the  water.     What  is  the  distance  of  the  ship  ? 

Ans.  21 1  miles. 

25.  The  distance  to  which  a  person  can  see  the  smooth 
surface  of  the  ocean,  if  no  allowance  be  made  for  refraction, 
is  equal  to  a  tangent  to  the  earth  drawn  from  his  eye,  as  T'A 
(Fig.  13.) 

Ex.  If  a  man  standing  on  the  level  of  the  ocean,  has  his 
eye  raised  5|  feet  above  the  water  ;  to  what  distance  can  he 
see  the  surface  ?  Ans.  2J  miles. 

*  See  Note  A. 


HEIGHTS  AND  DISTANCES.  11 

^iG.  If  the  distance  AT,  (Fig.  12.)  with  the  diameter  of  the 
earth  be  given,  and  the  height  BT  be  required  ;  the  equation 
in  art.  23  gives  

See  Surveying,  Section  IV.  on  Levelling, 
27.  When  the  diameter  of  the  earth  is  ascertained,  this 
may  be  made  a  base  line  for  determining  the  distances  of  the 
heavenly  bodies.  A  right  angled  triangle  may  be  formed,  the 
perpendicular  sides  of  which  shall  be  the  distance  required, 
and  the  semi-diameter  of  the  earth,  if  then  one  of  the  an- 
gles be  found  by  observation,  the  required  side  may  be  easily 
calculated. 

Let  AC  (Fig.  14.)  be  the  semi-diameter  of  the  earth.  All 
the  sensible  horizon  at  A,  and  CM  the  rational  horizon  paral- 
lel to  AH,  passing  through  the  moon  M.  The  angle  HAM 
may  be  found  by  astronomical  observation.  This  angle, 
which  is  called  the  Horizontal  Parallax^  is  equal  to  AMC, 
the  angle  at  the  moon  subtended  by  the  isemi-diameterof  the 
earth.     (Euc.  29.  1.) 

PttOBLEil  X. 

To  PIND  TJIE  DISTANCE  OF  ANY  HEAVENLY  BODY  WHOSE 
HORIZONTAL  PAKALLAX  IS   KNOWN. 

20.    As  RADIUS,  TO  THE   SEMI-DIAMETER  OP  THE   EARTH  J 

So  IS  THE  COTANGENT  OF  THE  HORIZONTAL  PARALLAX, 
TO  THE  DISTANCE. 

In  the  right  angled  triangle  ACM,  (Fig.  14.)  if  AC  be 
made  radius  ; 

R:  AC: :  Cot  AMC:  CM 

Ex.  If  the  horizontal  parallax  of  the  moon  be  0®  57',  and 
the  diameter  of  the  earth  7940  miles  ;  what  is  the  distance 
of  the  moon  from  the  centre  of  the  earth  ? 

Ans.  239,414  miles. 

29.  The  fixed  alars  are  too  far  distant  to  have  any  sensible 
horizontal  parallax.  But  from  late  observations  it  would 
seem,  that  some  of  them  are  near  enough,  to  suffer  a  small 
apparent  change  of  place,  from  the  revolution  of  the  earth 
round  the  sun.  The  distance  of  the  sun,  then,  which  is  the 
semi-diameter  of  the  earth's  orbit,  may  be  taken  as  a  base 
line,  (or  finding  the  distance  of  these  stars. 


12  MENSURATION  OF 

We  thus  proceed  hy  degree?,  from  measurin:;  a  line  on  the 
surface  of  the  earth,  to  calculate  the  distances  of  the  heaven- 
ly bodies.  From  a  base  line  on  a  plane,  is  determined  the 
height  of  a  mountain  ;  from  the  height  of  the  mountain,  the 
diameter  of  the  earth  ;  fronrj  the  diameter  of  the  earth,  the 
distance  of  the  sun,  and  from  the  distance  of  the  sun,  the  dis- 
tance of  Ih.  stars. 

30.  After  finding  the  distance  of  a  heavenly  body,  its  mag" 
nitude  is  easily  ascertained  ;  if  it  have  an  apj)arc  nt  diameter, 
sutliriently  iart^e  to  be  measured  by  the  instruments  which 
are  used  for  taking  angles. 

Let  AEB  (Fig.  15.)  be  the  angle  which  a  heavenly  body 
subtends  at  the  eye.  Half  this  angle,  if  C  be  the  centre  of 
the  body,  is  AEC ;  the  line  EA  is  a  tangent  to  the  surface, 
and  therefore  EAC  is  a  right  angle.  Then  making  the  dis- 
tance EC  radius, 

R:EC::SinAEC  :  AC 

That  is,  radius  Is  to  the  distance,  as  the  sine  of  half  the  an- 
gle which  the  body  subtends,  to  its  semi-diameter. 

Ex.  If  the  sun  subtends  an  angle  of  32'  2",  and  if  his  dis- 
tance from  the  earth  be  95  milhon  miles  ;  what  is  his  diam- 
eter ?  Ans.  885  thousand  miles. 


Promiscuous  Examples. 

1.  On  the  bank  of  a  river,  the  angle  of  elevation  of  a  tree 
on  the  opposite  side  is  found  to  be  46°  ;  and  at  another  sta- 
tion 100  feet  directly  back  on  the  same  level,  Sl*^.  What 
is  the  height  of  the  tree  ? 

Ans.  143  feet. 

2.  On  a  horizontal  plane,  observations  were  taken  of  a 
tower  standing  on  the  top  of  a  hill.  At  one  station,  the  an- 
gle of  elevation  of  the  top  of  the  tower  was  found  to  be  50°  ; 
that  of  the  bottom  39°  ;  and  at  another  station  150  feet  di- 
rectly back,  the  angle  of  elevation  of  the  top  of  the  tower 
was  3:2^.     What  are  the  heights  of  the  hill  and  the  tower? 

Ans.   The  hill  is  134  icci  high  ;  the  tower  G3. 

3.  What  is  the  altitude  of  the  sun,  when  the  shadow  of  a 
tree,  cast  on  a  horizontal  plane,  is  to  the  height  of  the  tree 
as  4  to  3?  Ans.  36°  52' 12" 


HEIGHTS  AND  DISTANCES.  1 8 

4.  If  a  straigV't  line  from  the  top  of  the  White  Mountains 
in  New-Hampshire  touch  the  ocean  at  the  distance  of  103^, 
miles  ;  what  is  the  height  of  the  mountains  ? 

Ans.  7100  feet. 

5.  From  the  top  of  a  perpendicular  rock  55  yards  high, 
the  angle  of  depression  of  the  nearest  bank  of  a  river  is  found 
(o  be  55''  54',  that  of  the  opposite  bank  33°  20'.  Required 
the  breadth  of  the  river,  and  the  distance  of  its  nearest  bank 
from  the  bottom  of  the  rock. 

The  breadth  of  the  river  is  46.4  yards ; 
Its  distance  from  the  rock  37.2 
C).  If  the  moon  subtend  an  angle  of  31'  14",  when  her  dis- 
tance is  240,000  miles  ;  what  is  her  diameter  ? 

Ans.  2180  miles. 

7.  Observations  are  made  on  the  altitude  of  a  balloon,  by 
two  persons  standing  on  the  same  side  of  the  balloon,  and  in 
a  vertical  plane  passing  through  it.  The  distance  of  the  sta- 
tions is  half  a  mile.  At  one,  the  angle  of  elevation  is  30°  58', 
at  the  other  36°  52'.  What  is  the  height  of  the  balloon 
above  the  ground  ?  Ans.  Ij  mile. 

8.  The  shadow  of  the  top  of  a  mountain,  when  the  alti- 
tude of  the  sun  on  the  meridian  is  32°,  strikes  a  certain  point 
on  a  level  plain  below  ;  but  when  the  meridian  altitude  of 
the  sun  is  67°,  the  shadow  strikes  half  a  mile  farther  south, 
on  the  same  plain.  What  is  the  height  of  the  mountain  above 
^be  plain.  Ans.  2245  feet. 


NAVIGATION. 


SECTION  I, 


Plane  Sailing. 


Art  '^'i  ]V"AV1GATI0N  is  the  art  of  conducting  a  ship 
-^^  on  the  ocean.  The  most  accurate  method  of 
ascertaining  ^he  situation  of  a  vessel  at  sea  is  to  find,  by  as- 
tronomical observations,  her  latitude  and  longitude.  But  this 
requires  a  view  of  the  heavenly  bodies  ;  and  these  are  often 
obscured  by  intervening  clouds.  The  mariner  must  there- 
fore have  recourse  to  other  means  for  determining  the  pro- 
gress which  he  has  made,  and  the  particular  part  of  the  ocean 
through  which  he  is  at  any  time  making  his  way.  The  com- 
mon method  is  to  measure  the  rate  of  the  ship's  going  by  a 
log-line,  and  to  find  the  direction  in  which  she  sails  by  a  ma- 
Ti?ier^s  compass.  From  these  data,  the  difference  of  latitude, 
the  departure,  and  the  difference  of  longitude,  may  be  calcu- 
lated. The  two  first  may  be  found  by  plane  sailing;  the  last 
by  middle  latitude  sailing,  or  more  correctly  by  Mercator's 
saihng.     See  Sec.  II.  and  III. 

34.  The  log-line  is  a  cord  which  is  wound  round  a  reel, 
one  end  being  attached  to  a  piece  of  wood  called  a  log.  It 
is  used  to  determine  the  distance  which  a  ship  runs  in  an 
bour,  by  measuring  the  distance  which  she  runs  in  half  a 
minute.  The  log  is  commonly  a  small  piece  of  board,  in  the 
form  of  a  quadrant  of  a  circle.  The  arc  is  loaded  with  a  quan- 
tity of  lead  sufficient  to  give  the  board  a  perpendicular  posi- 
tion, when  thrown  upon  the  water.  This  will  prevent  it 
from  moving  forward  toward  the  vessel,  while  the  line  is 
running  off  the  reel.     So  that  the  length  of  line  drawn  off 


16  NAVIGATION. 

by  the  log  in  half  a  minute,  is  equal  to  the  distance  which  the 
vessel  moves  through  the  water  in  that  time. 

The  log-line,  which  is  a  hundred  fathoms  or  more,  is  divi- 
ded into  equal  portions  called  knots.  Each  of  these  has  the 
same  ratio  to  a  nautical  mile,  which  half  a  minute  has  to  an 
hour.  That  is,  a  knot  is  the  1 20th  part  of  a  mile.  If  there- 
fore the  motion  of  the  ship  is  uniform,  she  sails  as  many  miles 
in  an  hour,  as  she  does  knots  in  half  a  minute. 

The  time  is  measured  by  a  half-minute  glass,  constructed 
like  an  hour  glass.  This  is  turned  when  the  log  is  thrown 
upon  the  water  ;  and  the  knots  drawn  from  the  reel,  while  the 
sands  are  running,  give  the  rate  of  the  ship.  The  log  is 
thrown  either  every  hour,  or  once  in  two  hours. 

35.  The  Mariner's  compass  is  a  circular  card,  attached  to 
a  magnetic  needle,  which  is  balanced  on  an  upright  pin,  so 
as  to  move  freely  in  any  direction.  The  ends  of  the  needle 
turn  towards  the  northern  and  southern  points  of  the  horizon. 
It  places  itself  in  the  magnetic  meridian,  which  nearly  coin- 
cides with  the  astronomical  meridian,  or  a  north  and  south 
line.*  Directly  over  the  needle,  a  line  is  drawn  on  a  card, 
one  end  of  which  is  marked  N,  and  the  other  S.  The  whole 
circumference  is  divided  into  equal  parts  by  32 points.  Four 
of  these,  the  N,  S,  E,  and  W,  are  called  cardinal  points. 
The  interval  between  two  adjacent  points  is  11°  15',  which 
is  the  quotient  of  360°  divided  by  32.  The  card  and  the 
needle  are  inclosed  in  a  circular  box,  on  the  inside  of  which 
a  black  markis  drawn  perpendicular  to  the  horizon.  When 
the  compass  is  placed  in  the  vessel,  a  line  passing  from  this 
mark  through  the  centre  of  the  card  should  be  parallel  to  the 
keel.  The  part  of  the  circumference  which  coincides  with 
the  mark  will  then  shew  the  point  of  compass  to  which  the 
keel  is  diiected.  To  prevent  the  needle  from  being  aflfect- 
ed  by  the  motion  of  the  vessel,  the  box,  and  a  brass  ring  by 
which  it  is  surrounded,  have  four  points  of  suspension  so  con- 
trived as  to  keep  the  card  nearly  parallel  to  the  horizon. 

"^  For  the  Variation  of  the  needle,  see  SrnvETiKC,  Sec.  V. 


PLANE  SAILING. 


IT 


The  following  is  a  table  of  the  number  of  degrees  and  min- 
utes corresponding  to  each  point  and  quarter  point  ol  the 
compass.     See  Fig.  16. 


Norm  Rist 
Uuailninf. 


jXorth. 
NjE 
NjE 
N|E 


N6E 
N6E1E 
NiEiE 
N6E|E 


NNE 
NNEiE 
NNEiE 
NNEfE 


NE^N 
NEfxN 

NEiN 

NEiN 


NE 
NEiE 
NEiE 
NE|E 


^outh-East 
Quatirant 


Soutk. 

SiE 
SfE 


Points 


0  0 
0  i 
0  i 

0   f 


S6E 
S6E1E 
S6E{E 
S6E5E 


1  0 

1  JL 

'  2 

1  # 


SSE 
SSE^E 
SSE^E 
SSEfE 


SE6S 
SEfS 
SEiS 
SEiS 


SE 

SEjE 
SEiE 
SE|E 


NE6E 
NE6EiE 
NE6EiE 
NE6E^E 


SE6E 
SE6EiE 
SE6E1E 
SEfeEfE 


2  0 

2  i 


^7  3 

3  0 

3  I 

3  i 

3  I 


4   0 

4   i 
4   f 


ENE 

1E6N|N 
E6NiN 
E6NiN 


EbN 
E|N 
EiN 
EJN 

East. 


ESE 

E6SfS 
E6S1S 
E6S1S 


EhS 
EfS 
EiS 
EiS 

E(ul. 


5   0 

5  I 


6  0 

6  i 

6  i 

6  I 


D   M. 


0  0  0 
2  48  45 
5  37  30 
8  2G  15 


11  15  0 

14  3  45 

IG  52  3() 

19  41  15 


22  JO  0 
25  18  4 
28  7  30 
30  5e   15 


33  45  0 
3G  33  45 
39  22  30 
42  11  15 


45  0  0 
47  48  45 
50  37  30 
53  26  15 


56  15  0 
59  3  45 
61  52  30 
64  41  15 


67  30  0 

70  18  45 

73  7  30 

75  56  15 


Soutti-West 
Quadrant 


South. 
SiVV 

s|w 


S6W 
S6W1VV 
S6WiVV 
S6W|W 


SSVV 

SSWiW 

SSVViVV 

ssw^w 


SVV6S 
SW|S 
SVViS 

swls 


sw 

SWiW 
SWiW 

SVVfVV 


SW6W 
SVV6VVi\V 
SVVAWiVV 
SW6W|W 


vvsw 

W6SfS 
VV6S1S 
W6SiS 


North- West 
Quadrant. 


Nurth. 
NiVV 

n|vv 


xNAVV 
N6VV|W 

N6VV1VV 


NNW 
NNVViW 
NNWiVV 
NNVV^VV 


NVV6N 

NW|N 
NVViiN 
NWiN 


iN  VV 

NWiW 
NVViW 
NWfVV 


45 
33 
22  30 
11    15 


0IW6S 
45 


90     0     0 


WfS 
WIS 
WIS 


Wesl. 


NVV^VV 
NVV6WiVV 
NVV/,VViVV 
NW6\V|W 


VViNVV 

W^NfN 

W/,NiN 


W6N 
WIN 
WIN 
VViN 
West. 


18  NAVIGATIOM. 

36.  Plane  Sailing  is  the  method  of  calculating  tlie  situ- 
ation and  piooicss  of  a  ship  hy  nneans  of  a  plane  triangle. 
Though  the  surface  of  the  ocean,  conforming  to  the  general 
figure  of  the  earth,  is  nearly  spherical  ;*  yet  the  quantities 
which  are  the  ohjects  of  inquiry  in  plane  sailing,  have  the 
same  relation?  to  each  other,  as  the  sides  and  angles  of  a  rec- 
tilinear triangle.  The  particulars  which  are  either  given  or 
recjuired  are/owr,  viz. 

1.  The  Course, 

2.  The  Distance, 

3.  The  Difference  of  Latitude, 
A,  The  Departure. 

37.  The  Course  is  the  angle  between  a  meridian  line  pass- 
ing through  the  ship,  and  the  direction  in  which  she  sails.  It 
is  described  by  saying  that  it  is  so  many  points  or  degrees 
cast  or  west  from  a  north  or  south  line.  Thus  if  the  vessel 
steers  NE  by  E,  the  course  is  said  to  be  N  5  points  E,  or  N 
56°15'E:  if  SSW,  it  is  said  to  be  S  2  points  W,  or  S 
22^°  W. 

A  ship  is  said  to  continue  on  the  same  course^  when  she 
cuts  ev<Ty  meridian  which  she  crosses  at  the  same  angle. 
She  is  steered  in  any  required  direction,  by  causing  the  keel 
to  make  a  constant  angle  with  the  needle.  The  line  thus  de- 
scribed is  not  a  straight  line,  nor  an  arc  of  a  circle,  but  a  pe- 
culiar kind  of  curve  called  the  Loxodromk\  spiral  ov  Rhumb- 
line. 

3!3.  The  Distance  is  the  length  of  the  line  which  the  ves- 
sel describes  in  the  given  time. 

39.  Difference  of  Latitude  is  the  distance  between  two  par- 
allels of  latitude,  measured  on  a  meridian.  It  is  also  called 
.Worthing  or  Southing. 

40  Departure  is  the  deviation  of  a  ship  east  or  west  from  a 
vieridian.  If  she  sails  on  a  parallel  of  latitude,  her  departure 
is  the  length  of  that  portion  of  the  parallel  over  which  she 
passes.     But  if  her  course  is  oblique,  she  is  continually  chan- 

*  The  true  figure  of  (he  earth  ia  nearer  a  spheroid  than  a  sphere.  But  the 
difference  is  too  inconsl<lerable  to  be  taken  into  account  in  any  calculations 
tor  which  the  lines  and  angles  are  given  from  the  log  and  the  compass.  In 
this  and  the  followin^^  seclions,  therefore,  the  earth  will  be  considered  as  a 
sphere 

+  From  Ao^cf  and  tJ'fC'-tof,  an  oblique  coil ree. 


PLANE  SAILING  .  19 

ging  her  latitude  ;  and  her  departure  for  each  instant  ought  to 
be  considered  as  measured  on  the  parallel  which  she  is  then 
crossing.  The  measure  will  not  be  correct,  if  it  be  taken 
wholly  on  the  parallel  wliich  the  ship  has  left,  or  on  that  upon 
which  she  has  arrived.  Suppose  she  proceed  from  A  to  C. 
(Fig.  18  )  Let  the  whole  distance  be  divided  into  indefinite- 
ly small  portions  Am,  mn,  nC  Draw  the  meridians  PM, 
PM  .  PM' ,  PM'"  ;  and  the  parallels  AD,  am.  y?i,  BC.  The 
departure  for  the  first  portion  is  om,  for  the  second  sn,  for 
the  third  ^C.  And  the  whole  departure  is  or)i-\-sn-\-tC; 
which,  on  account  of  the  obliquity  of  the  meridians,  is  less 
than  Bv-{-vt-}-tC  =  BC  the  meridian  distance  measured  on 
the  parallel  upon  which  the  ship  has  airived,  but  s;renter 
than  AD  the  meridian  distance  on  the  parallel  which  she  has 
left. 

41.  The  distance,  departure,  and  difference  of  latitude. 
are  measured  in  geographicnl  miles  or  minutes  ;  one  of  which 
is  equal  to  the  60th  part  of  a  degree  at  the  equator.  As  the 
circumference  o(  the  earth  is  about  25  thousand  English 
miles,  a  degree  is  nearly  69^  miles.  So  that  a  geographical 
or  nautical  mile  is  nearly  ^  greater  than  the  common  English 
toile.     A  league  is  three  miles. 

42.  The  peculiar  nature  of  the  Rhumb-line  gives  this  im- 
portant advantage  in  calculation,  that  the  distance,  departure, 
and  difference  of  latitude,  though  they  are  curve  lines,  may 
be  exactly  given  in  length  by  the  sides  of  a  riirJit-ang/ed  platie 
triangle,  in  which  one  of  the  angles  is  equal  to  the  course. 
Suppose  a  ship  proceeds  from  A  to  C,  (Fig.  18  )  describing 
the  rhumb-line  A/nnC,  on  which  the  angles  MAw,  M'mw, 
M"/iC  are  equal.  Let  the  whole  distance  be  divided  into 
portions  so  small,  that  the  triangles  kmo,  mns,  nCtf  shall  not 
differ  sensibly  from  plane  triangles.  The  meridians  and  par- 
allels being  drawn,  the  several  differences  of  latitude  are 
Ao.  m*,  nt ;  and  the  departures  ow,  sn,  tC,     (Art.  40.) 

In  the  straight  line  A'C  (Fig.  19  )  make  A'm'=  Am,  (Fig. 
18.)  ?n'n'=mn^  n'C'=nC,  and  the  angle  C'A'B'  =  mAo. 
Draw  m'v'  and  n't'  parallel  to  A'B' ;  and  m'o',  n'^',  and  C'B' 
perpendicular  to  A'B'.  Then  the  triangles  Amo,  \'>.i'o'.  mns, 
m'n's',  Ctn  and  C't'n'  are  all  similar  to^A'B'C.  The  differ- 
ence of  latitude  is 

AB  =  Ao-\-ms-{-nt  =  A'o'-hm's'  +  n't'  =  A'B'. 

And  the  departure  is 

om-{'sn+lC  =  o'm''\-s'n''{-f'C'=B'C', 


20  NAVIGATION. 

43.  In  plane  sailing,  then,  the  process  of  calculation  is  as 
accurate.,*  and  as  simple,  as  if  the  surface  of  the  ocean  were 
a  plane.  Let  NS  (Fig  20.)  be  a  meridian  line.  If  a  ship 
sails  from  A  to  C,  and  if  BC  is  perpendicular  to  NS;  then 

The  Course,  is  the  angle  at  A;  and  tlie  complement  of  the 
course,  the  angle  at  C  ; 

The  Distance  is  the  hypothenuse  AC  ; 

The  Depar  ure  is  the  base  BC,  which  is  always  opposite 
to  the  course;  and 

The  Difference  of  Latitude  is  the  perpendicular  AB,  which 
is  opposite  to  the  complement  of  the  course. 

Of  these  four  quantities,  any  two  heing  given,  the  others 
may  be  found  by  rectangular  trigonometry.  (Trig.  116.) 
The  parts  given  may  be 

1.  The  course  and  distance;  or 

2.  The  course  and  departure;  or 

3.  The  course  and  difference  of  latitude;  or 

4.  The  distance  and  departure; 

5.  The  distance  and  difference  of  latitude;  or 

6.  The  departure  and  difference  of  latitude. 

The  solutions  may  be  made  by  arithmetical  computation, 
by  Gunter's  scale  or  sliding  rule^  or  by  geometrical  construc- 
tion. (Trig  Sec.  Ill,  V,  VI.)  The  first  method  is  by  far 
the  most  accurate.  As  the  student  is  supposed  to  be  already 
familiar  with  trigonometry,  these  operations  will  not  be  repeat- 
ed here.  In  the  geometrical  construction,  it  will  be  proper 
to  consider  the  upper  side  of  the  paper  as  north,  and  the  low- 
er side  south.  The  right  hand  will  then  be  east,  and  the  left 
hand  west. 

Case  I. 

^^   ^.        S  The  course,      >^     ^   j  (  The  departure  and 
44.  G>ven  |  ^^^  dis,a„ce;  ]  '°  ""^  {  Difference  of  latitude. 

Here  we  have  the  hypothenuse  and  angles  given,  to  find 
the  base  and  perpendicular.      (Trig.  134.) 

Making  then  the  distance  radius, 
•D  J  .  T^-  . . .  ^  Sin    Course  :  Departure 
Rad.Dist..  I  j.^^  Course  :Diff.Lat. 

Example,    I . 
A  ship  sails  from  A  (Fig.  20.)  SW  by  S,  38  miles  to  C= 
Required  her  departure  and  difference  of  latitude  ?. 

♦See  Note  B. 


PLANE  SAILING.  21 

The  course  is  3  point*;,  or  33°  45    (Art.  35.) 
R  .  OQ. .  5  Sin    33°  4a'  :  21.1  =Drpjrl. 
^*^^'    I  Cos.  33°  45'  :  31.6=Diff.  Lat. 

Example  2. 

A  ship  sails  S  20°  E,  34  leagues.     Her  departure  and  dif- 
ference of  latitude  are  required. 

Ans.  16.5  and  29.7  leagues. 

The  proportions  in  this  and  the  following  cases  may  be  va- 
ried, by  making  different  sides  radius,  as  in  Trigonometrv 

Sec.  in. 

Case  II. 

Ar  r*'        (  The  course,       ")        /.    ,   (  The  distance,  and 
^^•^^'^"Und  departure;  i   '°  ^°^   ^  Difference  of  latitude. 
Making  the  distance  radius,  Trig.  137.) 
c     r>  .  T\  .  ^  Rad  :  Distance 

Sin  Course  :  Depart. .  ^  f,^^  bourse  :  Diff.  Lat. 

Example   1 . 
A  ship  leaving  a  port  in  latitude  42°  N,  has  sailed  S  3?° 
W,  till  she  finds  her  departure  62  miles.     What  distance  ha? 
she  run,  and  in  what  latitude  has  she  arrived .'' 
c-    Q-70  »  acy  ^  Rad  :  103=Distance. 
bm  J7    .  bJ. .  ^  ^^g  3^,  .  82.3==Diff.  Lat 

The  difference  of  latitude  is  S2.3  miles,  or  1°  I2'.3.  (Art. 
41.^  This  is  to  be  subtracted  from  the  original  latitude  of  the 
ship,  because  her  course  was  towards  the  equator.  The 
remainder  is  40°  47'. 7,  the  latitude  on  which  she  has  ar- 
rived. 

Example  2. 

A  ship  leaves  a  port  in  latitude  63°  S,  and  runs  N  54°  E, 
till  she  makes  a  harbour  where  her  departure  is  found  to  be 
74  miles  ;  how  great  is  the  distance  of  the  two  places,  and 
what  is  the  latitude  of  the  latter.'* 

The  distance  is91i  miles;  and  the  latitude  of  the  latter 
place  is  62°  06'.2. 

Case  III. 

4fi    r*         ^  ^^^  course,  and  ^  ♦    ^  j  ^  The  distance, 
(  Diff.  of  latitude  ;  J  (.  And  departure. 

Making  the  distance  radius, 

Cos  Course  :  Diff.  Lat: :  ^  ^."V  ^'Tf   ,r,  « 

(  Sm  Course  :  Departure. 


22  NAVIGATION. 

Example. 

A  ship  sails  S  50^  E,  from  latitude  T"  N,  to  latitude  4^  S. 
Required  her  distance  and  departure. 

As  the  two  latitudes  are  on  different  sides  of  the  equator, 
the  distance  of  the  parallels  is  evidently  equal  to  the  sum  of 
the  given  latitudes.  This  is  1 1°  or  660  miles.  The  distance 
is  1026.8  miles,  and  the  departure  786|. 

Case  IV. 

A'f   rirr  «  ^  The  distance,      1  ^    ^  a   S  The  course,  and 

^^•^'^^"^  And  departure;  ]  ^°  ^"^   1  Diff.  of  latitude. 

Making  the  distance  radius,  (Trig.  135.) 

Dist  :  Rad: :  Depart  J  Sin  Course, 

Rad  :  Dist:  :Cos  Course  :  Diff.  Lat. 

Example. 

A  ship  having  left  a  port  in  Lat.  3°  N,  and  sailing  between 
S  and  E  400  miles,  finds  her  departure  ISO  miles.  What 
course  has  she  steered,  and  what  is  her  latitude? 

Her  latitude  is  2°  57'i  S,  and  her  course  S  26°  44'J  E. 

Case  V. 

AQ    n-  ^  The  distance,  and  >   ^    n    ,   ^  The  course, 

48.  Given.  <  ri/r     r  i  .-.  j         t   to  find   w     ,   ,         / 

(  Ditt.  ol  latitude;    )  ^  And  departure. 

Making  the  distance  radius, 
Dist  :  Rad: : Diff.  Lat  :  Cos  Course, 
Rad  :  Dist:  :Sin  Course  :  Departure. 

Example, 

A  vessel  sails  between  N  and  E  66  miles,  from  Lat.  34°  50 
to  Lat.  35°  40'.     Required  her  course  and  departure. 

The  course  is  N  40°  45'  E,  and  the  departure  43.08  miles. 

Case  VL 

49,  Given.  \  ^^^  ^7^!-^^'  '"^  \  '^  ^"d    \  f  ^^^  ^.^"^^^' 

^  Diff.  of  latitude  ;      5  ^  And  distance. 

Making  the  difference  of  latitude  radius,  (Trig,  139.) 
Diff.  Lat  :  Rad:  :Depart  :  Tan  Course, 
Rad  :  Diff.  Lat:  :Sec  Course  :  Distance. 

Example. 
A  ship  sails  from  the  equator  between  S  and  W,  till  her 


TRAVERSE  TABLE.  23 

latitude  is  5^  52',  and  her  departure  264  miles.     Required 
her  course  and  distance. 

The  course  is  S  36°  52'i  W,  and  the  distance  is  440  miles. 

Examples  for  practice, 

1.  Given  a  ship's  course  S  46°  E,  and  departure  59  miles; 
10  find  the  distance  and  difference  of  latitude. 

2.  Given  the  distance  68  miles,  and  departure  47;  to  find 
the  course  and  difference  of  latitude. 

3.  Given  the  course  SSE,  and  the  distance  57  leagues;  to 
find  the  departure  and  difference  of  latitude. 

4.  Given  the  course  NW  by  N,  and  the  difference  of  lati- 
tude 2°  36' ;  to  find  the  distance  and  departure. 

5.  Given  the  departure  92,  and  the  difference  of  latitude 
86  :  to  find  the  course  and  distance. 

6.  Given  the  distance  123,  and  the  difference  of  latitude 
07  ;  to  find  the  course  and  departure. 


THE  TRAVERSE  TABLE. 


50.  To  save  the  labour  of  calculation,  tables  have  been 
prepared,  in  which  are  given  the  departure  and  difference  of 
latitude,  for  every  degree  of  the  quadrant,  or  for  every  quar- 
ter of  a  degree.  These  are  called  Traverse  tables,  or  tables 
oi  Departure  and  Latitude.  The  distance  is  placed  in  the 
left  hand  column,  the  departure  and  difference  of  latitude  di- 
rectly opposite,  and  the  degrees  if  above  45°  or  4  points,  at 
the  top  of  the  page,  but  if  under  45°,  at  the  bottom.  The 
titles  at  the  top  of  the  columns  correspond  to  the  courses  at 
the  top  ;  and  the  titles  at  the  bottom,  to  the  courses  at  the  bot- 
tom ;  the  difference  of  latitude  for  a  course  ^rea/er  than  45°, 
being  the  same  as  the  departure  for  one  which  is  as  much 
/c55  than  45°.     See  Trig.  104. 

If  the  given  distance  is  greater  than  any  contained  in  the 
table,  it  may  be  divided  into  parts,  and  the  departure  and  dif- 
ference of  latitude  found  for  each  of  the  parts.  The  sums  of 
the  numbers  thus  found  will  be  the  numbers  required. 


24  NAVIGATION. 

The  departure  and  difference  of  latitude  for  decimal  parts 
may  be  found  in  the  same  manner  as  for  whole  numbers,  by 
supposing  the  decimal  point  in  each  of  the  columns  to  be 
moved  to  the  left,  as  the  case  requires. 

With  the  aid  of  a  traverse  table,  all  the  cases  of  plane  sail- 
ing may  be  easily  solved  by  inspection. 

Ex.  1.  Given  the  course  33°  45';  and  the  distance  38 
miles;  to  find  the  departure  and  difference  of  latitude. 

Under  SJ*^!,  and  opposite  38,  will  be  found  the  difference 
of  latitude  31.6,  and  the  departure  21.11 ;  the  same  as  in 
page  21. 

2.   Given  the  course  57°,  and  the  distance  163. 

The  departure  and  diff.  of  lat.  for  100  are  83.87  and  54.46 

for    63        52.84         34.31 


for  163      136.71         88.77 


3.   Given  the  course  39°,  and  the  distance   18.23. 
The  departure  and  diff.  of  lat.  for  18.    are  11.33  and  13.99 

for      .23        0.14  0.18 


for  18.23      11.47         14.17 


4.  Given  the  course  41°  15',  and  the  departure  60. 
Under  41°^,  and  against  the  departure  60.  will  be  found 

the  difference  of  latitude  68.42  and  the  distance  91. 

5.  Given  the  distance  63,  and  the  departure  56. 
Opposite  the   distance  63,  find  the  departure  56 ;  in  the 

adjoining  column   will  be  the  latitude  28.85,  and  at  the  bot- 
tom, the  course  62°|. 

6.  Given  the  departure  72,  and  the  difference  of  latitude 
37. 

Opposite  these  numbers  in  the  columns  of  latitude  and  de- 
parture, will  be  found  the  distance  81,  and  at  the  foot  of  the 
columns,  the  course  62°f . 

51.  The  traverse  table  is  useful,  not  only  for  taking  out 
departure  and  difference  of  latitude;  but  for  finding  by  in- 
spection the  sides  and  angles  of  any  right-angled  triangle 
whatever  In  plane  sailing,  the  distance  is  the  hypothenuse, 
(see  Fig.  20.)  the  difference  of  latitude  is  the  perpendicular, 
the  departure  is  the  base,  and  the  course  is  the  acute  angle 


PARALLEL  SAILING.  25 

at  the  perpendicular.  If  then  the  hypotlienusc  of  any  right- 
afigled  triangle  whatever,  be  found  in  the  column  of  distances, 
in  the  traverse  table;  the  perpendicular  will  be  opposite  in 
the  latitude  column,  and  the  base  in  the  departure  column; 
the  angle  at  the  perpendicular,  being  at  the  top  or  bottom  of 
the  page. 

Ex.  1.  Given  the  hypothenuse  24,  and  the  angle  at  the 
perpendicular  54°A  ;  to  find  the  base  and  perpendicular  by 
inspection. 

Opposite  24  in  the  distance  column,  and  over  54°|  will  be 
found  the  base  19.54  in  the  departure  column,  and  the  per- 
pendicular 13.94  in  the  latitude  column. 

2.  Given  the  angle  at  the  perpendicular  37°|,  and  the  base 
46  ;  to  find  the  hypothenuse  and  perpendicular. 

Under  37°i.  look  for  46  in  the  departure  column  ;  and 
opposite  this  will  be  found  the  perpendicular  60.5  in  the  lati- 
tude column,  and  the  hypothenuse  76  in  the  distance  col- 
umn. 

3.  Given  the  perpendicular  36,  and  the  base  30.21  ;  to 
find  the  hypothenuse  and  angles. 

Look  in  the  columns  of  latitude  and  departure,  till  the  num- 
bers 38  and  30.21  are  found  opposite  each  other ;  these  will 
give  the  hypothenuse  47,  and  the  angle  at  the  perpendicular 
40°. 


SECTION  II. 

PARALLEL  AND  MIDDLE  LATITUDE  SAILING. 


52.  By  the  methods  of  calculation  in  plane  sailing,  a  ship's 
course,  distance,  departure,  and  difference  of  latitude  are  found. 
There  is  one  other  particular  which  it  is  very  important  to 
determine,  the  difference  of  longitude.  The  departure  gives 
the  distance  between  two  meridians  in  miles.  But  the  situa- 
tions of  places  on  the  earth,  are  known  from  their  latitudes 
and  longitudes;  and  these  are  measured  in  degrees.  The 
lines  of  longitude,  as  they  are  drawn  on  the  globe,  are  farthest 

5 


26  NAVIGATION. 

from  each  other  at  the  equator,  and  gradually  converge  to- 
wards the  poles.  A  ship,  in  making  a  hundred  miles  of 
departure,  may  change  her  latitude  in  one  case  2  degrees,  in 
another  iO,  and  in  another  20.  It  is  important,  then,  to  be 
able  to  convert  departure  into  difference  of  longitude  ;  that  is, 
to  determine  how  many  degrees  of  longitude  answer  to  any 
given  number  of  miles,  on  any  parallel  of  latitude.  This  is 
easily  done  by  the  following 

Theokem. 

53.    As  THE  cosine  of  latitude, 
To  radius  ; 

So    IS   THK  DEPARTURE, 
To    THE    DIFFERENCE    OF    LONGITUDE. 

By  this  is  to  be  understod.  that  the  cosine  of  the  latitude  is 
to  radius;  as  the  distance  between  two  meridians  measured 
on  the  given  parallel,  to  the  distance  between  the  same  me- 
ridians measured  on  the  equator. 

Let  P  (Fig.  21.)  be  the  pole  of  the  earth,  A  a  point  at  the 
equator,  L  a  place  whose  latitude  is  given,  and  LO  a  line  per- 
pendicular to  PC.  Then  CL  or  CA  is  a  semi-diameter  of 
the  earth,  which  may  be  assumed  as  the  radius  of  the  tables; 
PL  is  the  complement  of  the  latitude,  and  OL  the  sine  ofPL, 
that  is,  the  cosine  of  the  latitude. 

If  the  whole  be  now  supposed  to  revolve  about  PC  as  an 
axis,  the  radius  CA  will  describe  the  equator,  and  OL  the 
given  parallel  of  laiitude.  The  circumferences  of  these  cir- 
cles are  as  their  semi  diameters  OL  and  CA,  (Sup  Euc.  8. 
1.)  And  this  is  the  ratio  which  any  portion  of  one  circum- 
ference has  to  a  like  portion  of  the  other.  Therefore  OL  is 
to  CA,  that  is,  the  cosine  of  latitude  is  to  radius,  as  the 
distance  between  two  meridians  measured  on  the  given  par- 
allel, to  the  distance  between  the  same  meridians  measured 
on  the  equator. 

Cor.  1.  Like  portions  of  different  parallels  of  latitude  are 
to  each  other,  as  the  cosines  of  the  latitudes. 

Cor.  2.  A  degree  of  longitude  is  commonly  measured  on 
the  equator.  But  if  it  be  considered  as  measured  on  a  paral- 
lel of  latitude,  the  lc?igth  of  the  degree  will  he  as  the  cosine  of 

the  latitude. 


PARALLEL  SAILING. 


The  following  table  cojitains  the  length  of  a  decree  oj  longi 
tilde  fur  each  degree  of  latitude. 


D.t 

-UiUs 

I)  1, 

Alil^ 

J)  1- 

.•Vlile, 

D.L 

.Miles 

D  1, 

Milts 

j.i. 

.MileK.     1 

\   l'59.99j 

|16 

57.67  |31i51.43!46 

41.68  ;6li29.09| 

76 

14.52; 

!  2 

5jI.96 

Il7 

57.3S 

32  50.88;  47 

40.92 

62 

28.17 

77 

13.50| 

3 

59.92 

18 

57.06 

;  33  50.32'  48 

40.15 

,63 

27,24 

78 

12.47; 

4 

59.85 

1^^,56  73||34|49.74| 

49 

39.36 

!64 

26  30 

79 

U.45i 

i^ 

59.77 

20  56.38 

35 

36 

49.15 
48.54 

bO 

38.57 

j65 

25.36 
24.40 

SO 

8I1 

10.42J 

1 

9  39i 

6 

59.67; 

21  56  Oi 

51 

37.76 

l66 

1  7 

50  55 

22 

55-63 

37 

47  92 

5'2|36.94|';67 

23.44 

82 

8.35| 

'  6 

59  42 

23 

55.23 

38  47.28 

oi 

36.11 

68 

22.45 

!83| 

7.31 

9 

59.26] 

24 

54.81 

39  46.63 

54 

.35.27 

69 

2/50 

84i 

6.27 

10 

59.08 

25 

54  38 

1" 

41 

45.96 

3.5  34.41 1 

70 

20.52 

85 

5.23J 

11 

58.S9I26 

53.93 

43.28 

06 

33.55 

71 

19.53 

861!  4.19 

12 

58.68 

27 

53.4rJ 

42  44.59||57 

32.68 

72 

18  54 

:87 

3.141 

13i58.46 

2Si52.97| 

43  43.88 

58 

31.80 

73 

17.54 

'88; 

2.091 

14jo6.22  29  52  47! 

44  43.16 

59 

30-90 

174  16.54 
75I15.53 

89 i   1.05 

15157  95!'30  51.96; 

45  42.43 

60 

30.00 

90  1  0,00; 

The  length  of  a  degree  of  longitude  in  different  parallels  is 
also  shown  by  the  Line  of  Longitude,  placed  over  or  under 
the  line  of  chords,  on  the  plane  scale.     See  Trig.  (165.) 

The  sailing  of  a  ship  on  a  parallel  of  latitude*  is  called  Par- 
allel Sailing.  In  this  case,  the  departure  is  equal  to  the  dis- 
tance. The  difference  of  longitude  may  be  found  by  the 
preceding  theorem;  or  if  the  difference  of  longitude  be  giv- 
en, the  departure  may  be  found  by  inverting  the  terms  of  the 
proportion.     (Alg.  380.  3.) 

55.  The  Geometrical  Construction  is  very  simpld.  Make 
CBD  (Fig.  22.)  a  right  angle,  draw  BC  equal  to  the  depar- 
ture in  miles,  lay  off  the  angle  at  C  equal  to  the  latitude  in 
degrees,  and  draw  the  hypothenuse  CD  for  the  difference  of 
longitude.  The  angles  C,  and  the  sides  BC  and  CD,  of  this 
triangle,  have  the  same  relations  to  each  other,  as  the  latitude, 
departure,  and  difference  of  longitude. 

For  Cos  C  :  BC:  :R  :  CD    (Trig.  121.) 

And  Cos  Lat  :  Depart:  :R  ;  Diff.  Lon.  (Art.  53. 


?ee  Note  C^ 


28  NAVIGATION. 

5Q.  The  parls  of  the  triangle  may  be  found  by  inspection 
in  the  traverse  table.  (Art.  51.)  The  angle  opposite  the 
departure  is  D  the  complement  ol  the  latitude,  and  the  differ- 
ence of  longitude  is  the  hypothenuse  CD.  If  then  the  de- 
parture be  found  in  the  departure  column  under  or  over  the 
given  number  of  degrees  in  the  co-latitude,  the  difference  of 
longitude  will  be  opposite  in  the  distance  column. 

Example  L 

A  ship  leaving  a  port  in  Lat.  38°  N.  Lon.  1G°  E.  sails  west 
on  a  parallel  of  latitude  117  miles  in  24  hours.  What  is  her 
longitude  at  the  end  of  this  time  .'* 

Cos38°  :  Rad::ll7  :  148'^=2°  28'^    the    difference  of 

longitude. 
This  subtracted  from  1G°  leaves  13°  31'^  the  longitude  re- 
quired. 

Example  11. 

What  is  the  distance  of  two  places  in  Lat.  46°  N.  if  the 
longitude  of  the  one  is  2°  13'  W.  and  that  of  the  other  1° 
17' E.? 

As  the  two  places  are  on  opposite  sides  of  the  first  meridian, 
the  difference  of  longitude  is  2°  13'  + 1°  17'  =  3°  30',  or  210 
minutes.     Then 

Rad  :  Cos  46°:  :210  :  145.88  miles,  the  departure,  or  the 
distance  between  the  two  places. 

Example  IIL 

A  ship  having  sailed  on  a  parallel  of  latitude  138  miles, 
finds  her  difference  of  longitude  4°  3'  or  243  minutes.  What 
is  ht  r  latitude? 

Diff.  Lon.  243  :  Dep.  138! : Rad  :  Cos  Lat.  55°  23'f. 

Example  IV. 

On  what  part  of  the  earth  are  the  degrees  of  longitude  half 
a's  long  as  ftie  equator  ? 

AnSr    In  latitude  60. 


MIDDLE  LATITUDE  SAILLXG.  09 


Middle  Latitude  Sailing. 

o7.  By  the  method  just  explained,  is  calculated  the  dilTer- 
ence  of  longitude  of  a  ship  sailing  on  a  parallel  of  latitude. 
But  instances  of  this  mode  of  sailing  are  comparatively  few. 
It  is  necessary  then  to  be  able  to  calculate  the  longitude  when 
the  course  is  oblique.  If  a  ship  sail  from  A  to  C,  (Fig.  lo.) 
the  departure  is  ecjual  to  om-\-sn-\-tC,  But  the  sum  of 
these  small  lines  is  /tS5  than  BC,  and  greater  than  AD.  (Art. 
40.)  The  departure,  then,  is  the  meridian  distance  measur- 
ed not  on  the  parallel  from  which  the  ship  sailed,  nor  on 
that  upon  which  she  has  arrived,  but  upon  one  which  is  be- 
tween the  two.  If  the  exact  situation  of  this  intermediate 
parallel  could  be  determined,  by  a  process  sufficiently  sim- 
ple for  common  practice,  the  difference  of  longitude  would 
be  easily  obtained.  The  parallel  usually  taken  for  this  pur- 
pose, is  an  arithmetical  mean  between  the  two  extreme  lati- 
tudes. This  is  called  the  Middle  Latitude,  The  meridian 
distance  on  this  parallel  is  not  exactly  equal  to  the  depar- 
ture. But  for  small  distances,  the  errour  is  notmateiial,  ex- 
cept in  high  latitudes. 

The  middle  latitude  is  equal  to  half  the  sum  of  the  two  ex- 
treme latitudes,  if  they  are  both  north  or  both  south  :  but 
to  half  their  difference. y  if  one  is  north  and  the  other  south. 

58.  In  middle  latitude  sailing,  all  the  calculations  are 
made  in  the  same  way  as  in  plane  sailing,  excepting  the  pro- 
portions in  which  the  difference  of  longitude  is  one  of  the 
terms.  The  departure  is  derived  from  the  difference  of  lon- 
gitude, and  the  difference  of  longitude  from  the  departure, 
in  the  same  manner  as  in  parallel  sailing,  (Arts.  53,  54.)  only 
substituting  in  the  theorem  the  term  middle  latitude  for  lat. 
itude. 

Theorem  I. 

As  THE  COSINE  OF  MIDDLE  LATITUDE^ 

To  RADIUS   ; 

So  IS  THE  DEPARTURE, 

To  THE  DIFFERENCE  OF  LONGITUDE. 

59.  The  learner  will  be  very  much  assisted  in  stating  the 
proportions,  by  keeping  the  geometrical  construction  steadi- 
ly in  his  mind.  In  Fig.  20  we  have  the  lines  and  angles  in 
plane  sailing,  and  in  Fig.  22,  those  in  parallel  sailing.     By 


30  NAVIGATION.. 

briiif^'ing  these  together,  as  in  Fig.  23,  we  have  all  the  parts 
in  middle  latitude  sailing.  The  two  right  angled  triangles, 
being  united  at  the  common  side  BC,  which  is  thci  departure, 
Ibrm  the  oblique  angled  triangle  ACD. 

60.  The  angle  at  D  is  the  complement  of  the  middle  lati- 
tude.    (Art.  55.)     Then  in  the  triangle  ACD,  (Trig.  143.) 
Sin  D  :  AC:  :Sin  A  :  DC  ;  that  is, 

Theorem  II. 

As  THE  COSINE  OF  MIDDLE  LATITUDE, 

To  THE  DISTANCE  ; 

So  IS  THE  SINE  OF  THE  COURSE, 

To  THE  DIFFERENCE  OF  LONGITUDE. 

Gl.  The  two  preceding  theorems,  with  the  proportions  in 
plane  sailing,  are  sufficient  for  solving  all  the  ci^ses  in  middle 
latitude  sailing.  A  third  may  be  added,  for  the  sake  of  re- 
ducing two  proportions  to  one. 

In  the  triangle  BCD  (Fig.  23.)  Cos  BCD  :  R:  :BC  :  CD 
And  in  the  triangle  ABC,  AB  :  R : :  BC  :  Tan  A. 

The  means  being  the  same  in  these  two   proportions,  the 

extremes   are  reciprocally  proportional.     (Alg.  387.)     We 

have  then 

Cos  BCD  :  AB:  :Tan  A  :  CD;  that  is, 

Theorem  III. 

As  the  cosine  of  middle  latitude. 
To  the  difference  of  latitude  ; 
So  is  the  tangent  of  the  course, 
To  the  difference  of  longitude. 

Among  the  other  data  in  middle  latitude  sailing,  one  of  the 
extreme  latitudes  must  always  be  given. 

Example  L 

At  what  distance,  and  in  what  direction,  is  Montock  Point 
iVom  Martha's  Vineyard  ;  the  former  being  in  Lat.  4l°  Ol'N. 
Lon.  72°  W.,  and  the  latter  in  Lat.  41°  17'  N.  Lon.  70°  48' 
W.? 

Here  are  given  the  two  latitudes  and  longitudes,  to  find 
the  course  and  distance. 

The  difference  of  longitude  is     72' 
The  difference  of  latitude  13' 

The  middle  latitude  41°  10'\ 


lUlDDLE  LATITUDE  SAILING.  3i 

Beginning  with  the  triangle  in  which  there  are  two  parts 
given,  by  theorem  I, 

R  :  Diff.  Lon:  :Cos  Mid.  Lat  ;  Depart.=54.2 

And  by  plane  sailing,  Case  Vf, 
DiflT.  Lat  :  Rad:  ".Depart  :  Tan  Course  =  7G°  30'| 
Or  to   tind    the  course  at  a  single  statement,  by  theorem 

ni, 

Diff.  Lat  :  Cos  Mid.  Lat:  :Diff.  Lon  :  Tan  Course  =  7G°  30'| 
To  find  the  distance  by  plane  sailing.  Case  III, 
Cos  Course  :  Diff.  Lat:  :Rad  :  Dist.  =  55.73 

Example  IL 

A  ship  leaving  New-York  light-house  in  Lat  40^^  28'  N. 
and  Lon.  74°  08'  W.  sails  S.  E.  67  miles  in  24  hours.  Re- 
quired her  latitude  and  longitude  at  the  end  of  that  time. 

By  plane  sailing, 
Rad  :  Dist:  :Cos  Course  :  Diff.  Lat.=47'.4 
The  latitude  required,  therefore,  is  39°  40'. 6,  and  the  mid- 
dle latitude  40°  04'.3. 

Then  by  theorem  II, 
Cos  Mid.  Lat  :  Dist:  :Sin  Course  :  Diff.  Lon.=61'.9 

Or  by  theorem  III, 

Cos  Mid.    Lat :  Diff.   Lat: : Tan  Course  :  Diff.  Lon.=Gl'.9 

The  longitude  required  is  73°  06'.  1. 

Example  III, 
A  ship   leaving  a  port  in  Lat   49°  57'  N.  Lon.  5°  14'  W. 
sails  S.  39°  W.  till  her   latitude  is  45°  31'.     Required  her 
longitude  and  distance. 

Ans.  10°  34'.3  W.  and  342.3  miles. 

Example  IV, 

A  ship  sailing  from  Lat.  49°  57'  N.  and  Lon.  5°  14' W. 
steers  west  of  south,  till  her  longitude  is  23°  43',  and  her  de- 
parture 789  miles.  Required  her  course,  distance,  and  lat- 
itude. 

Course       51°  5' W. 

Latitude     39°  20'  N. 

Distance    10(4  miles. 


SECTION  III. 

MERCATOR'S  SAILING. 

A  T  62  nn^^  calculations  in  middle  latitude  sailing  are 
'  '  JL.  simple,  and  sufficiently  accurate  for  short  dis- 
tances, particularly  near  the  equator.  But  they  become 
quite  erroneous,  when  applied  to  great  distance,  and  to  high 
latitudes.  The  only  method  in  common  use,  which  is  strict- 
ly accurate,  is  that  called  Mcrcator''s  Sailings  or  Wright's 
Sailioi/.  This  is  founded  on  the  construction  of  a  chart,  pub- 
lished in  1556  by  Gerard  Mercator.  About  forty  years  after, 
Mr.  Edwnrd  Wright  gave  demonstrations  of  the  principles  of 
this  chart,  and  applied  them  to  the  solution  of  problems  in 
navigation. 

63.  In  the  construction  of  Mercator's  chart,  the  earth  is 
supposed  to  be  a  sphere.  Yet  the  meridians,  instead  of  con- 
verging towards  the  poles,  as  they  do  on  the  globe,  are  drawn 
parallel  to  each  other.  The  distance  of  the  meridians,  there- 
fore, is  every  where  too  great,  except  at  the  equator.  To 
compensate  this»  the  degrees  of  latitude  are  proportionally 
enlarged.  On  :he  artificial  globe,  the  parallels  of  latitude 
are  drawn  at  equal  distances.  But  on  Mercator's  chart,  the 
distances  of  the  parallels  increase  from  the  equator  to  the 
poles,  so  as  every  where  to  have  the  same  ratio  to  the  dis- 
tances of  the  meridians,  which  they  have  on  the  globe. 
Thus  in  latitude  60%  where  the  distance  of  the  meridians 
must  be  doubled,  to  make  it  the  same  as  at  the  equator,  a  de- 
gree of  latitude  is  also  made  twice  as  great  as  at  the  equa- 
tor. The  dimensions  of  places  are  extended  in  the  projec- 
tion, in  proportion  as  they  are  nearer  the  poles.  The  diam- 
eter of  an  island  in  latitude  60°  would  be  represented  twice 
as  great  as  if  it  were  on  the  equator,  and  its  area  four  times 
as  great. 

♦  Robertson's  Navigation,  London  Phil  Trans,  for  1666  and  1696,  Hut- 
ton's  Dictionary,  Introduction  to  Hutton's  Mathematical  Tables,  Bowditch's 
Practical  Navigator,  Emerson'?  an^  M'Laurin's  Fluxions,  M'Kay'?  Naviga- 
tion, Emerson's  Pria.  Navig,  Barrow's  Navigation. 


MERCATOR'S  SAILING.  33 

64.  Table  of  Meridional  Parts,  If  a  meridian  on  a  sphere 
be  divided  into  degrees  or  minutes,  the  portions  are  all  equal. 
But  in  Mercator's  projection,  they  are  extended  more  and 
more  as  they  are  farther  from  the  equator.  To  facilitate  the 
calculations  in  navigation,  tables  have  been  prepared,  which 
contain  the  length  of  any  number  of  degrees  and  minutes  on 
this  extended  meridian,  or  the  distance  of  any  point  of  the 
projection  from  the  equator.  These  are  called  tables  of  Me- 
ridional Parts.  The  common  method  of  computing  them 
is  derived  from  the  following  proposition. 

65.  Any  MixuTf:  portion  of  a  parallel  of  latitude, 
Is  to  a  like  portion  of  the  meridian  ; 

As  radius, 

To  THE   SECANT  OF  THE  LATITUDE. 

For.  by  the  theorem  in  parallel  sailing,  (Art.  53.)  the  co- 
sine of  latitude  is  to  radius,  as  the  departure  to  the  difference 
of  longitude  measured  on  the  equator  ;  that  is,  as  a  part  of 
the  parallel  of  latitude,  to  a  like  part  of  the  equator.  But  on 
a  sphere,  the  equator  and  meridian  are  equal. 

Therefore  Cos  Lat  :  Rad:  '.a  part  of  the  parallel:  a  like 
part  of  the  meridian. 

But  Cos  Lat  :  Rad: : Rad  :  Sec  Lat.  (Trig.  93.  3.) 

By  equality  of  ratios  then,  (Alg.  384.) 
Apart  oj  the  parallel  :  a  like  part  of  the  merid: !  Rad  :  Sec  Lat. 

By  like  parts  of  the  parallel  of  latitude  and  the  merid- 
ian are  here  meant  minutes,  seconds  or  other  portions  of  a 
degree.  The  proposition  is  true  when  applied  either  to  the 
circles  on  a  sphere,  or  to  the  lines  in  Mercator's  projection. 
For  the  parts  of  the  latter  have  the  same  ratio  to  each  other, 
as  the  parts  of  the  former.  (Art.  63.)  The  divisions  of 
Mercator's  meridian,  however,  should  be  made  very  small; 
for  the  measure  of  each  part  is  supposed  to  be  taken  at  the 
parallel  of  latitude,  and  not  at  a  distance  from  it.  In  the  com- 
mon tables,  the  meridian  is  divided  into  minutes. 

66.  Suppose  then  that  the  length  of  each  minute  of  a  de- 
gree of  Mercator's  meridian  is  required.  By  the  proposition 
in  the  last  article, 

1'  of  the  parallel  I  I'  of  the  meridian.:  iRad  :  Sec.  Lat. 

But  in  this  projection,  the  parallels  of  latitude  are  all  equal. 

6 


M  NAVIGATION. 

(Art.  63.)  Wliaiever  be  tne  latitude,  then,  the  first  term  of 
the  proportion  is  equal  to  a  minute  at  the  equator,  or  a  geo- 
graphical mile;  and  if  this  is  assumed  as  the  radius  of  the 
trigonometrical  tables  (  Irig.  100.)  the  first  and  third  terms 
are  equal,  and  therefore  the  second  and  fourth  must  be  equal 
also.  (Alg.  695,)  That  is,  the  length  of  ayiij  one  minute  of 
jMercatur's  mertdmn  is  equal  to  the  natural  secant  of  the  lati- 
tndi  of  that  part  of  the  meridian, 

Vu^-^"'    ,  ?   minme  of  ihe  meridian  is  (  f  ^ '"i""'e. 
1  he  second  \  ,  ,^    ,,,^  ^^^^^^  ^^    /   u>«  mmutes, 

Ihe  third     )       ^  (^  mrec  mmutes, 

he.  &ic. 

The  table  of  meridional  parts  is  formed  by  adding  together 
the  several  minutes  thus  found.*  Beginning  from  the  equa- 
tor, an  arc  of  the  meridian 

o(  two    minutes=st'C   \'-\-sec  2', 
of  three  n\\nu\es= sec   1  -\-sec  2'-{-sec  3', 
of  four  minutes=5ec  I'-j-^ec  2'-^sec  3'-f  sec  4', 
he.  &ic. 
See  the  table  at  the  end  of  this  number. 
To  find  from  the  table  the  length  of  any  given  number  of 
degrees  and  minutes,  look  for  the  degrees  at  the  top  of  the 
page,  and  the  minutes  on  the  side  ;  then  against  the  minutes, 
and  under  the  degrees,  will  be  the  length  of  the  arc  in  nau- 
tical miles. 

67.  Meridional  Difference  of  Latitude,  An  arc  of  Mer- 
cator's  meridian  contained  between  tzoo  parallels  of  latitude^ 
is  called  meridional  difference  of  latitude.  It  is  found  by 
subtractini^  the  meridional  parts  for  the  less  latitude  from  the 
meridional  parts  for  the  greater,  if  both  are  north  or  south ; 
or  by  adding  them,  if  one  latitude  is  north  and  the  other 
south. 
Thus  the  lat.  of  Boston  is  42°  23'  Merid.  parts  2S13 
Baltimore  39    23  Merid.  parts     2575 


Proper  difFerence  of  lat.       3°         Merid.  diff.  of  lat.  238 

68.  If  one  latitude  and  the  meridional  difFerence  of  lati- 
tude be  given^  the  proper  difFerence  of  latitude  is  found  by 
reversing  this  process. 

*  See  Note  D. 


MERCATOR'S  SAILING.  35 

When  the  two  latitudes  are  on  the  same  side  of  the 
equator,  subiiaciiiig  the  niendional  difference  of  latitude 
from  the  meridional  parts  for  the  greater,  will  give  the  me- 
ridional parts  for  the  less;  or  adding  the  meridional  ditier- 
ence  to  the  parts  for  the  less  latitude,  will  give  the  parts 
for  the  greater  But  il  the  two  latitudes  are  on  opposite 
sides  of  the  equator,  subtracting  the  parts  lor  the  one  lat* 
rtude  from  the  meridional  difference,  will  give  the  parts  for 
the  other. 

Thus  the  meridional  difference  of  latitude  between 

New- York  and  New-Orleans  is  793 

The  lat.  of  N.  Orleans  is  29°  57  Merid.  parts       1SS5 

The  lat  of  New-York       40   42  Merid.  parts       267S 

69.  Sulutio7is  171  Mercator'*s  Sailing.  The  Solutions  in 
Mercator's  sailing  are  founded  on  t/ie  swiUaritu  of  tic o  right- 
angled  triangles,  in  one  of  7chich  the  perpendicular  sides  arc 
the  proper  difference  of  I  at  dude  and  the  departure  ;  and  in  the 
other  the  r  eridional  dfference  of  latitude  and  the  difference  of 
longitude. 

According  to  the  principle  of  Mercator's  projection,  the 
enlargement  of  each  minute  portion  of  the  meridian  is  pro- 
portioned to  the  enlargement  of  the  parallel  of  latitude  which 
crosses  it.  (Art.  63.)  Any  part  of  the  meridian  before  it  is 
enlarged,  is  proper  dfference  of  latitude  j  and  after  it  is  en- 
larged, is  meridional  difference  of  latitude,  A  part  of  the 
parallel,  before  it  is  enlarged,  is  departure^  and  after  it  is  en- 
larged, is  equal  to  the  corresponding  difference  of  longitude  ; 
because  in  this  projection,  the  distance  of  the  meridians  is 
the  same  on  any  parallel,  as  at  the  equator,  where  longitude 
is  reckoned. 

If  then  we  take  a  small  portion  of  the  distance  which  a  ship 
has  sailed,  as  A?/i,  (Fig.  18.) 

Prop.  Dif  Lat.  Ao  ;  Depart,  om: : Merid.  Dif  Lat  i  Dif.  Lon. 
In  the  triangle  ABC,  (Fig.  24.)  let  the  angle  at  A  = 
the  course  oA//?,  (Fig.  18.)  AB=the  proper  difference 
of  latitude.  AC  =  Am4-'"'i-h"C  the  distance,  and  BC  = 
om-{-sn-rtC  the  departure.  Then  as  the  triangles  Aoniy 
msn,  7itC  are  each  similar  to  the  trian2;!e  ABC,  (Fig, 
24.)  the  difference  of  latitude  for  any  one  of  the  small  dis- 
tances as  Am,  is  to  the  corresponding  departure ;  as  the 
whole  difference  of  latitude  AB  to  the  zrholc  departure  BC. 
Therefore. 


3t>  NAVIGATION. 

(  for  Am  :  i  for  Am, 

P.  Dif.  Lat.  AB  :  Dep.  BC  : :  Mer.  Dif.  Lat.  )  for  mn  :  Dif.  Lon.    )  for  mn^ 

(  for  nC  :  (for  nC. 

But  the  whole  meridional  difference  of  latitude  for  the  dis- 
tance AC,  is  equal  to  the  sum  of  the  differences  for  A??;,  mn, 
and  7iC;  and  the  whole  difference  of  longitude  is  equal  to  the 
sum  of  the  differences  for  Am,  mn,  and  nC.  Therefore,  (Alg. 
388.  Cor.  1.) 
Prop.  Dif.  Lat.  AB  :  Dep.  BC:  :Merid.  Dif.  Lat  :  Dif.  Lon. 

Extend  AB,  (Fig.  24.)  making  Al  equal  to  the  meridional 
difference  of  latitude  corresponding  to  the  proper  difference 
of  latitude  AB  ;  from  L  draw  a  line  parallel  to  BC,  and  ex- 
tend AC  to  intersect  this  in  D  Then  is  DL  the  differenct 
of  longitude.  For  it  has  been  shown  that  the  difference  of 
longitude  is  ^fourth  prdportionul  to  thej  proper  difference  of 
latitude,  the  departure,  and  the  meridional  difference  of  lati* 
tude  ;  and  by  similar  triangles, 

AB  :  BC::AL  :  LD. 

70.  To  solve  all  the  cases,  then,  in  Mercator's  sailing,  we 
have  only  to  represent  the  several  quantities  by  the  parts  of 
two  similar  right-angled  triangles,  as  ABC  and  ALD,  (Fig. 
24.)  and  to  find  their  sides  and  angles.  In  the  smaller  trian- 
gle ABC  the  parts  are  the  same  as  in  plane  sailing,  and  the 
calculations  are  made  in  the  same  manner.  The  sides  AL 
and  DL  are  added  for  finding  the  difference  of  longitude;  or 
when  the  difference  of  longitude  is  given,  to  derive  from  it 
one  of  the  other  quantities.  The  course  is  common  to  both 
the  triangles,  and  the  complement  of  the  course  is  either 
ACB  or  ADL.  The  hypothenuse  AD  is  not  one  of  the 
quantities  which  are  given  or  required  in  navigation. 

71.  In  the  similar  triangles  ABC,  ALD,  (Fig.  24.) 

AB  :  AL::BC  :  LD;  that  is, 

Theorem  I. 

As  THE  PROPER  DIFFERENCE  OF  LATlTtJDE, 

To  THE  MERIDIONAL  DIFFERENCE  OF  LATITUDE  ; 

So  IS  THE  DEPARTURE, 

To  THE  DIFFERENCE  OF  LONGITUDE. 

72.  In  the  triangle  ALD,  if  AL  be  made  radius. 

Bad  :  Tan  A::AL  :  DL;  that  is, 


MERCATOR'S  SAILING.  37 

Theorem  II. 

As    RADIUS, 

To  THE  TANGENT  OF    THE  COURSE  ; 

So  IS  THE  51KRIDIONAL  DIFFEUINCE  OF   LATITUDE, 

To  THE  DIFFERENCE  OF  LONGITUDE. 

By  this  theorem,  the  difference  of  longitude  may  be  calcu- 
lated, without  previously  finding  the  departure. 

73.  In  Mercator's  as  well  as  in  middle  latitude  sailing, 
one  latitude  mi  st  always  be  given.  This  is  requisite  in  con- 
verting proper  difference  of  latitude  and  meridional  difference 
of  latitude  into  each  other.     (Arts.  67,  68.) 

74.  When  the  difference  of  latitude  is  veiy  sniall^  the  dif- 
ference of  longitude  will  be  more  correctly  found  by  middle 
latitude  sailing,  than  by  Mercator's  sailing;  unless  a  table  is 
used  in  which  the  meridional  parts  are  given  to  decwials. 
Mercator's  sailing  is  strictly  correct  in  theory.  But  the  com- 
mon tables  are  not  carried  to  a  degree  of  exactness,  sufficient 
to  mark  very  minute  differences.  On  the  other  hand,  the 
errour  of  middle  latitude  sailing  is  diminished,  as  the  differ- 
ence of  latitude  is  lessened. 


c  41' 

The  latitudesofMontock  and  Martha's  Vineyard  are  <  .. 


Example  I. 

'      °4'N. 
17  N. 

Their  longitudes  <  ^^  ^g,  y^' 

Required  the  course  and  distance  from  one  to  the  other. 

Lat.  of  Martha's  Vin,  41°  17'       Merid.  parts  2724 
of  Montock  41  04       Merid.  parts  2707 

Proper  Diff.  of  Lat.  13'  Mer.  Dif.  Lat.       17  (Art.  67.) 

The  difference  of  longitude  is  1*  12'=72  miles. 

To  find  the  course  by  theorem  II,  (Fig   24.) 
Merid.  Diff.  Lat  :  Diff.  Lon::Rad  :  Tan  Course=76°  43.' 

To  find  the  distance  by  plane  sailing, 
Cos  Course  :  Prop.  Diff.  Lat.:  :Rad  :  Dist.=56.58. 

The  results  by  middle  latitude  sailing,  page  31,  are  a  littfe 
different,  as  that  method  is  not  perfectly  accurate. 


38  NAVIGATION. 

Example   IL 

A  ship  sailing  from  the  Lizard  in  Lat.  49®  57'  N.  Lon.  b'^ 
15'  W.  proceeds  S.  39°  W  till  her  latitude  is  found  by  obser- 
vation to  be  45°  31'  N.  What  is  then  her  longitude,  and 
what  distance  has  she  run  ? 

Here  are  given  the  difference  of  latitude  and  the  course,  to 
find  the  distance  and  tlie  difference  of  longitude. 

The  proper  difference  of  latitude  is  4°  26'=266' 

The  meridional  difference  of  latitude  396 

Then  by  plane  sailing, 

Cos  Course  :  Prop.  Diff.  Lat.: : Had  :  Dist.=342.3. 

And  by  theorem  II, 

Had  :  Tan  Course  ::M.Dif.  Lat  :  Uif  Lon.=320'.7=5°2(y.T 

This  added  to  the  longitude  of  the  Lizard  5°  14'  gives  the 
longitude  of  the  ship  10°  34'.7  W. 

Example  IIL 
A  ship  sailing  from  Lat.  49°  57'  N.  and  Lon.  5°  14'  W. 
steers  west  of  south,  till    her  latitude   is  39°  20'  N.  and  her 
departure  789   miles.     Required   her  course,   distance,  and 
longitude. 

The  proper  difference  of  latitude  is  10°  37'=637' 

The  meridional  difference  of  latitude  899 

Then  by  theorem  I,  (Fig.  24  ) 

P.  Dif.  Lat.  :  M.  Diff.   Lat::Dep  :  Diff.  Lon.=:1113'.5^ 

18°33'.5 
The  longitude  of  the  ship  is  therefore  23°  47'^ 

And  by  plane  sailing, 
Prop  Diff.   Lat  :  Rad::Depart  :  Tan  Course=51°  5' 
Rad  :  Prop.  Diff.  Lat. :  :Sec  Course  :  Distance=1014  miles. 
Example  IV, 
A  ship  sailing  from  a  port  in  Lat.  14°  45'  N.  Lon.  17°  33' 
W.  steers  S.  28°  7'i   VV.  till  her  longitude  is  found   by  ob- 
servation to  be  29°  26'  W.  Required  her  distance  and  latitude. 

The  difference  of  longitude  is  11°  53'=713'. 
By  theorem  II, 
Tan  Course  :  Rad: : Dif.  Lon  :  M  Dif  Lat.  =  1334  S. 
Lat.  of  the  port  14°  45'  N.     Merid.  parts    895  N. 

of  the  ship    7  18  S.     Merid.  parts  439  S.  (Art.  68.) 


Diff.  of  Lat.     22°  3' =  1323' 


MERCATOR'S  SAILING.  39 

By   plane  sailing, 
Cos  Course  :  Diff.  Lat:  :Rad  :  Distance  =1500  miles. 

Example  V, 

A  ship  sails  300  miles  between  north  and  west,  from  Lat. 
37°  N.  to  41°  N.  What  is  her  course  and  difference  of  longi- 
tude ? 

The  course  is  N.  36°  52'  W.,  and  the  difference  of  longi- 
tude 3°  52!. 

Example   VI. 
A  ship  sails  S.  67°  30'  E.  from  Lat   50°  10'  S  till  her  de- 
parture is  957   miles.     What   is  her  distance,  difference  of 
latitude,  and  difference  of  longitude  ? 

The  distance  is  1030  miles. 

The  difference  of  latitude       G°  36'.4 
The  difference  of  longitude  26°  53' 
Example  VII. 
A  ship  sailing  from  Lat.  26°  13'  N.  proceeds  S.  27°  W. 
231  miles.     What  is  her  difference  of  latitude  and  difference 
of  longitude? 

Example  III, 
A  ship  sailing  from  Lat.  14°  S.  260  miles,  between  south 
and  west,    makes   her   departure    173   miles.     What  is  her 
cburse,  difference  of  latitude,  and  difference  of  longitude  r* 

♦  See  Note  E- 


SECTION  IV. 

TRAVERSE  SAILING. 

TJY  the  methods  in  the  preceding  sections,  are 
Art.  7j.  J-J  found  the  difference  of  latitude,  departure,  &;c. 
for  a  single  course.  But  it  is  not  often  the  fact  that  a  ship 
proceeds  from  one  port  to  another  in  a  direct  line.  Varia- 
ble and  contrary  winds  frequently  render  a  change  of  direc- 
tion necessary  every  few  hours.  The  irregular  path  of  the 
ship  sailing  in  this  manner,  is  called  a  traverse. 

Resolving  a  traverse  is  reducing  the  compound  course  to 
a  single  one.  This  is  commonly  done  at  sea  every  noon. 
From  the  several  courses  and  distances  in  the  log-book,  the 
departure,  difference  of  latitude,  &c.  are  determined  for  the 
whole  24  hours.  In  the  same  manner,  the  courses  of  several 
successive  days  are  reduced  to  one,  so  as  to  ascertain,  at  any 
lime,  the  situation  of  the  ship.  The  following  methods  by 
construction  and  by  calculation,  are  sufficiently  accurate  for 
short  distances,  at  least  near  the  equator. 

76.  Geometrical  construction  of  a  traverse.  To  construct 
a  traverse,  draw  a  meridian  line  and  lay  down  the  first  course 
and  distance  ;  from  the  end  of  this,  lay  down  the  second 
course  and  distance  ;  from  the  end  of  that,  a  third  course, 
&ic.  Then  draw  a  line  connecting  the  extremities  of  the  first 
and  last  of  these,  to  show  the  whole  distance,  and  the  direc- 
tion of  the  ship  from  the  point  of  starting. 

This  will  be  easily  understood  by  an  example. 
Example  L 

A  ship  sails  from  a  port  in  Lat.  32°  N.,  and  in  24  hours 
makes  the  following  courses  ; 

1.  N.  25°  E.  16  miles, 

2.  S.  54°  E.  11, 

3.  N.  13°  W     7, 

4.  N.  61°  E.     5, 

5.  N.  38°  W.  18, 

It  is  required  to  find  the  departure,  difference  of  latitude, 
distance,  and  course,  for  the  whole  traverse. 


TRAVERSE  SAILING.  41 

On  A  as  a  centre  (Fig.  25.)  describe  a  circle  and  draw  the 
meridian  NAS.  Then  considering  the  upper  part  as  north, 
the  right  hand  east,  and  the  left  hand  west,  draw  the  hnes 
Al,  A2,  A  J,  A4,  and  A5,  to  correspond  with  the  several 
eourses;  that  is,  make  the  angle  NA1=25°,  SA2=54°; 
NA^3=13°,  NA4=61°,  and  NA5=38^ 

Make  AlB=16,  BC  =  11  and  parallel  to  A2,  CD=7 
and  parallel  to  A3,  DF  =  5  and  parallel  to  A4,  FG=18 
and  parallel  to  A5  ;  join  AG,  and  draw  GP  perpendicular 
loNS. 

Then  if  the  surface  of  the  ocean  be  considered  as  a  plane, 
G  is  the  place  of  the  ship  at  the  end  of  the  24  hours,  AG  the 
distance  from  port,  PG  the  departure,  AP  the  difference  of 
laliiufh,  and  GAP  the  course.  The  angles  may  be  measured 
by  a  line  of  chords,  and  the  distances  taken  from  a  scale  of 
equal  parts.^     (^Trig.  148,  161,  2.) 

The  distance  is  32.3  miles. 

The  departure  7.38 

The  difference  of  lat.  31  45 

The  course  13^  12' 

77.  Resolving  a  traverse,  by  Calculation  or  Inspection, 
When  a  ship  sails  on  different  courses  for  a  short  time,  the 
difference  of  latitude,  at  the  end  of  that  time,  is  equal  to  the 
difference  between  the  sum  of  the  northings  and  the  sum  of 
ihe  southings,  and  the  departure  is  neaily  equal  to'the  differ- 
ence between  the  sum  of  the  eastings  and  the  sum  of  the 
westings.  See  Arts.  78  79,  If  then  the  difference  of  lati- 
tude and  the  departure  for  each  course  be  found  by  calcula- 
tion or  inspection,  and  placed  in  separate  columns  in  a  table; 
the  difference  of  latitude  for  the  whole  time  mav  be  obtained 
exactly,  and  the  departure  nearly,  by  addition  and  subtrac- 
tion ;  and  the  corresponding  distance  and  course  may  be  de- 
termined by  tri^^onometrical  calculation  or  inspection,  as  in 
the  last  case  of  plane  sailing.     (Art.  49.) 

The  following;  table  contains  the  courses,  distances,  depart- 
ure, and  difference  of  latitude  in  the  preceding  example.  See 
Fig.  25, 

•^  See  Note  F. 
7 


42 


NAVIGATION. 


Traverse  Table. 


Courses. 

Distances. 

w- 

Lai, 

Departure.    1 

N. 

S. 

.  .. . 

6.47 

E. 

6.76 
8.90 

4.37 

1 

1   w. 

I    N.  25°  E. 
2.  S.    54°  E. 
3    N.    13°  W 

4.  N.   61°  E 

5.  N.  38°  W. 

N   13°  12^  E. 

AB   16 
BC  11 
CD     7 
DF     5 
FG  18 

AG.  32.3 

14.50 

6.82 

2.42 

14  18 

1.57 
11.08 

37.92 
6.47 

6.47 

20  03 
12.65 

12.65 

1  31.45 

7.38 

The  sum  of  the  northings  is  37.92.  Subtracting  from  this 
the  southing  6.47,  we  have  the  difference  of  latitude  AP  31.45 
N. 

The  sum  of  the  eastings  is  20.03.  Subtracting  from  this 
the  sum  ot  the  westings  12.65  we  have  the  departure  GP 
7.38  E.     Then  (Art.  49.) 

Diff.  Lat  :  Rad::Depart  :  Tan  Course  NAG=13°  12^1 

Rad  :  Diff.  Lat: : Sec.  Course  :  Distance  AG =32.3 


The  latitude  of  the  port  is 
The  difference  of  latitude 

TheJatitude  of  the  ship 

The  meridional  difference  of  lat. 


32°  N. 
0°  31'.45  N. 

32°  31'.45  N. 
37.5 


Then  by  Mercator's  sailing, 
Rad  :  Tan  Course:  iMerid.  Diff.  Lat  :  Diff.  Lon.=8'.8 

Example  IL 

A  ship  sailing  from  a  port  in  Lat.  42°  N.  makes  the  follow- 
ing courses  and  distances. 


1.  S,   13°  E. 

21  miles, 

2.  S.   18°  W. 

16, 

3.  N.  84°  E. 

9, 

4.  S.  67°  E. 

12, 

5.  N.  78°  E. 

14, 

6.  S.  12°  W. 

35. 

TRAVERSE  SAILING.  43 

The  difference  of  latitude,  departure,  &ic.  are  required. 
The  departure  is  26'. 19  E. 

The  difF.  of  latitude,  P  lO'f  S. 

The  diff  of  longitude,  35'.07 

The  direct  course,  S.     20°  18'|  E. 
The  distance,  75^  miles. 

Accurate  method  of  resolving  a  traverse, 

78.  The  preceding  method  of  resolving  a  t'-averse  is  fre- 
quently used  at  sea,  because  it  is  simple,  and  in  most  cases 
is  sufficiently  accurate  for  a  run  of  24  hours.  But  it  is  found- 
ed on  the  assumption,  that  when  a  ship  sails  from  one  place 
to  another  by  several  courses,  she  makes  the  sarne  departure^ 
as  if  she  had  proceeded  by  :i  single  course  to  the  same  place. 
This  is  not  strictly  true.  Suppose  a  vessel,  instead  of  sail- 
ing directly  from  A  to  C,  (Fig  i8.)  proceeds  by  one  course 
from  A  to  H.  and  then  by  a  different  course  from  H  to  C. 
In  the  compound  course,  the  whole  departure,  is  6f/+^H  + 
tC  ;  (Art.  40.)  which  on  account  of  the  obliquity  of  the  me- 
ridians, is  less  than  om-\-sn-\~tC,  the  departure  on  the  single 
course.  If  the  compound  course  had  been  on  the  other  side 
of  the  single  one,  nearer  the  equator,  the  departure  would 
have  been  greater, 

71).  But  the  difference  of  latitude  is  the  same,  whether  the 
ship  proceeds  from  one  place  to  the  other,  on  a  single  course, 
or  on  several  The  difference  of  latitude  AB  (Fig.  18.)  = 
\o-\-ms-\-nt==Kb-\-dg-\-Y{t,  The  difference  of  longitude  is 
also  the  same,  whether  the  course  is  single  or  compound. 
For  the  difference  of  longitude  is  the  distance  between  the 
meridians  of  the  two  places  measured  on  the  equator. 

If  then  the  difference  of  latitude  and  difference  of  longitude 
be  calculated  for  each  part  of  the  compound  course  ;  the  whole 
difference  of  latitude  and  difference  of  longitude  icill^be  found 
by  addition  and  subtraction  ;  and  from  these  may  be  deter- 
mined the  direct  course  and  distance.  The  difference  of 
longitude  for  each  course  may  be  obtained  independently  of 
the  departure,  by  theorem  II.  of  Mercator's  sailing. 

It  will  facilitate  the  calculation  of  the  longitude,  to  place  in 
the  traverse  table,  the  latitudes  at  the  beginning  and  end  of 
each  of  the  courses,  the  corresponding  meridional  parts,  and 
the  meridional  differences  of  latitude. 


44 


NAVIGATION. 


In  the  followlnpj  example,  the  courses  and  distances  are 
the  sanne  as  in  Art.  7G.  Ex.  1.  The  port  from  which  the 
ship  is  supposed  to  sail,  is  in  latitude  32°  N. 


1 

Traverse  Table. 

Utg  Lut. 

Latitudes 

-  ^.  (n.a. 

Mend. 

Diff.  Long.i 

Courses. 

Dist 



Parts. 

Dtf.  Lai. 

N.     1      S. 

32° 

;12    14 '.50 

2028 
2045.5 

17.5 
7.5 

E        1      W. 

I.  N.  25°  E 

16 

14.50 

6.47 

8.16 

2.  S.  54°  E. 

11 

J2       8  03 

2038 

7.8 

10.32 

3.  N    13°  W. 

/ 

6.82 

32     14.85 

2045.8 

2  5 

1.80 

4    N.  61°  E 

5 

2.42 

32     17.27 

2048.3 

17.2 

4.51 

5.  N.38°  W. 

18 

14  1!; 

r37.92 
6.47 

647 

32    3145 

2065.5 

22.99 
15.24 

13.44 
15.24 

11°  40'  37" 

32.12i 

31.45 

' 

7.75 

The  difference  of  longitude  is  here  found  to  be  7.75,  and 
in  Art.  77,  8'. 8  the  errour  there  bein^  1.05. 

To  find  the  direct  course  and  distance  from  the  port  to  the 
place  of  the  ship. 

Merid.  Dif.  I.at  :  Dif.  Lon::Rad  :  Tan  Conrse  =  ll°40'  37" 
Rad  :  Prop.  Dif.  LatllSec  Course  :  Distance=32.12. 

By  comparing  the  results  here  with  those  in  Art.  77,  it  will 
be  seen  that  a  small  errour  was  introduced  there,  both  into 
the  course  and  the  distance^  by  making  them  dependent  on 
the  departure  ;  which  being  obtained  from  ihe  several  cour- 
ses, is  not  the  same  as  for  a  single  course.     (Art.  78.) 

Ex.  2  A  ship  sailing  from  a  port  in  latiude  78°  15'  N. 
fnakes  the  following  courses  and  distances. 


1. 

N. 

67°  3(y    w. 

154  miles. 

2. 

S. 

45             W. 

96 

3. 

N. 

50    371  W. 

89 

4. 

N. 

11     15     E. 

110 

5. 

N. 

36    33  J  W. 

56 

6. 

S. 

19    4U  E. 

78 

Required  the  difference  of  latitude,  the  difference  of  lon- 
gitude, and  the  distance  the  ship  must  have  sailed,  to  reach 
the  same  place  on  a  single  course. 

The  difference  of  latitude  is  2*  7' 
The  difference  of  longitude  22°  29' 
The  direct  course  N.  63°     1'  W. 

The  distance  .  279.9  miles. 


SECTION  V. 

MISCELLANEOUS  ARTICLES. 

I.  The  Plane  Chart, 

-^  f  SlHE  Charts  commonly  used  in  navigation  are  eitber 
J_  Plane  Charts,  or  Mercator's  Charts,  The  latter 
are  generally  to  be  preferred.  But  plane  charts  will  answer 
for  short  distances,  such  as  the  extent  of  a  harbour  or  small 
bay. 

In  the  construction  of  the  plane  chart,  that  part  of  the  sur- 
face of  the  globe  which  is  represented  on  it,  is  supposed  to 
be  a  plane.  The  meridians  are  drawn  parallel  ;  and  the 
lines  of  latitude  at  equal  distances.  Islands,  coasts,  &ic.  are 
delineated  upon  it,  by  laying  down  the  several  parts  accord- 
ing to  their  known  latitudes  and  longitudes. 

81.  On  a  chart  extending  a  small  distance,  each  side  of 
the  equator,  the  meridians  ought  to  be  at  the  same  distance 
from  each  other,  as  the  parallels  of  latitude.  A  similar 
construction  is  frequently  applied  to  different  parts  of  the 
globe.  But  this  renders  the  chart  much  more  incorrect 
than  is  riecessary.  A  circular  island  in  latitude  60  would, 
by  such  a  construction,  be  thrown  into  a  figure  whose  length 
from  east  to  west  would  be  twice  as  great,  as  from  north 
to  south ;  the  comparative  distance  of  the  meridians  being 
made  twice  as  great  as  it  ought  to  be.  (Art.  53.  Trig. 
96.   cor.) 

But  when  the  chart  extends  only  a  few  degrees,  if  the 
distance  of  the  meridians  is  proportioned  to  the  distance 
of  the  parallels  of  latitude,  as  the  cosine  of  the  mean  latitude 
to  radius;  (Art,  53.)  the  representation  will  not  be  mate- 
rially incorrect.  The  meridian  distance  in  the  middle  of  the 
chart  will  be  exact.  On  one  side,  it  will  be  a  little  too 
great;   and  on  the  other,  a  little  too  small. 

82.  To  construct  a  Plane  Chart,  then,  on  one  side  of  the 
paper  draw  a  scale  of  equal  parts,  which  are  to  be  counted 
as  degrees  or  minutes  of  latitude,  according  to  the  proposed 
x'xtent  of  the  chart.     Through  the  several  divisions,  draw  tht 


46  THE  PLANE  CHART. 

parallels  of  latitude,  and  at  right  angles  to  these,  draw  the 
meridians  in  such  a  manner,  that  their  distance  from  each 
other  shall  be  to  the  distance  of  the  parallels  of  latitude,  as 
the  cosine  of  the  latitude  of  the  middle  of  the  chart,  to  ra- 
dius. 

After  the  lines  on  all  the  sides  are  graduated,  the  po- 
sitions of  the  several  places  which  are  to  be  laid  down, 
may  be  determined,  by  applying  the  edge  of  a  rule  or  strip 
of  paper,  to  the  divisions  for  the  given  degree  of  longitude 
on  each  side,  and  another  to  the  divisions  for  the  degree 
of  latitude.  In  the  intersection  of  these,  will  be  the  point 
required. 

The  distance  which  a  ship  must  sail,  in  going  from 
one  place  to  another,  on  a  single  course,  may  be  nearly 
found,  by  applying  the  measure  of  the  interval  between  the 
two  places,  to  the  scale  of  miles,  of  latitude  on  the  side  of 
the  chart.* 

H.  Construction  of  Mercator's  Chart. 

83.  In  Mercator's  chart,  the  meridians  are  drewn  at  equal 
distances,  and  the  parallels  of  latitude  at  unequal  distances, 
proportioned  to  the  meridional  differences  of  latitude.  (Arts. 
63,  67.)  To  construct  this  chart,  then,  make  a  scale  of  equal 
parts  on  one  side  of  the  paper,  for  the  lowest  parallel  of  lati- 
tude which  is  to  be  laid  down,  and  divide  it  into  degrees  and 
minutes.  Perpendicular  to  this,  and  through  the  dividing 
points  for  degrees,  draw  the  lines  of  longitude.  For  the  se- 
cond proposed  parallel  of  latitude,  find  from  the  table,  (Art. 
67.)  the  meridional  difference  of  latitude  between  that  and 
the  parallel  first  laid  down,  and  take  this  number  of  minutes 
from  the  scale  on  the  chart,  for  the  interval  between  the  two 
parallels.  Iti  the  ssme  manner,  find  the  interval  between  the 
Second  and  third  parallels,  between  the  third  and  fourth,  &ic. 
till  the  projection  is  carried  to  a  sufficient  extent. 

Places  whose  latitudes  and  longitudes  are  known,  may  be 
laid  down  in  the  same  manner  as  on  the  plane  chart,  by  the 
intersection  of  the  meridians  and  lines  of  latitude  passing 
through  them. 

If  the  chart  is  upon  a  small  scale,  the  least  divisions  on  the 
graduated  lines  may  be  decrees  instead  of  minutes  ;  and  the 
meridians  and  parallels  may  be  drawn  for  every  fifth  or  every 
tenth  degree.     But  in  this  case,  it  will  be  necessary  to  di- 

^See  Note  G. 


MERCATOR'S  CHART.  47 

vide  the  meridional  differences  of  latitude  by  60,  to  reduce 
them  from  minutes  to  degrees. 

84.  The  Line  of  Meridional  Parts  on  Gunler'^s  scale  is  di- 
vided in  the  same  manner  as  Mercator's  Meridian,  and  cor- 
responds with  the  table  of  meridional  parts  ;  except  that  the 
numbers  in  the  latter  are  minutes^  while  the  divisions  on  the 
other  are  degrees.  Directly  beneath  the  line  of  meridional 
parts,  is  placed  a  line  of  equal  parts.  The  divisions  of  the 
latter  being  considered  as  degrees  of  longitude,  the  divisions  of 
the  former  will  be  degrees  of  latitude  adapted  to  the  same 
scale.  The  meridional  difference  of  latitude  is  found,  by  ex- 
tending the  compasses  from  one  latitude  to  the  other. 

A  chart  may  be  constructed  from  the  scale,  by  using  the 
line  of  equal  parts  for  the  degrees  of  longitude,  and  the  line 
of  meridional  parts  for  the  intervals  between  the  parallels  of 
latitude. 

85.  It  is  an  important  property  of  Mercator's  chart,  that 
all  the  rhumb-lines  projected  on  it  are  straight  lines.  This 
renders  ii,  in  several  respects,  more  useful  to  navigators,  than 
even  the  artificial  globe.  By  Mercator's  sailing,  theorem  II, 
(Art.  72.) 

Merid.  DifF.  Lat  :  Diff.  Lon!  :Rad  :  Tan  Course 

So  that,  while  the  course  remains  the  same,  the  ratio  of 
the  meridional  difference  of  latitude  to  the  difference  of  lon- 
gitude is  constant.  If  A,  C,  C',  and  C"  (Fig.  26.)  be  several 
points  in  a  rhumb-line,  AB  AB',  and  AB",  the  correspond- 
ing meridional  differences  of  latitude,  and  BC,  B'C,  B'  C", 
the  differences  of  longitude  j  then 

AB  :  BC::AB'  :  BC::AB"  :  B'Q'. 

Therefore  ABC,  AB'C,  and  AB"C".  are  similar  triangles^ 
and  ACC'C"  is  a  right  line.     (Euc.  32.  6.) 

III.     Obl:q,ue  Sailing. 

86.  The  application  of  oblique  angled  trigonometry  to  the 
solution  of  certain  problems  in  navigation,  is  called  oblique 
sailing.  It  is  principally  used  in  bays  and  harbours,  to  de- 
termine the  bearings  of  objects  on  shore,  with  their  distances 
from  the  ship  and  from  each  other.  A  few  examples  will 
be  sufficient  here,  in  addition  to  those  already  given  under 
heights  and  distances. 


ly  NAVIGATION. 

One  of  the  cases  which  most  frequently  occurs,  is  that  in 
which  the  distance  of  a  ship  from  land  is  to  be  determined, 
when  leaving  a  harbour  to  proceed  to  sea.  This  is  necessa- 
ry, that  her  difference  of  latitude  and  departure  may  be  reck- 
oned from  a  fixed  point,  whose  latitude  and  longitude  are 
known. 

The  distance  from  land  is  found,  by  taking  the  bearing  of 
an  object  from  the  ship,  then  running  a  certain  distance,  and 
taking  the  bearing  again.  The  course  being  observed,  there 
will  then  be  given  the  angles  and  one  side  of  a  triangle,  to 
find  either  of  the  remaining  sides. 

Example  L 

The  point  of  land  C  (Fig.  27.)  is  observed  to  bear  N  ^1^ 
30'  W.  from  A.  The  ship  then  sails  S.  67^  30'  W.  9  miles 
from  A  to  B;  and  the  direction  of  the  point  from  B  is  found 
to  be  N.  11°  15'  E.  At  what  distance  from  land  was  th« 
ship  at  A  ^ 

Let  NS  and  N'S'  be  meridians  passing  through  A  and  B. 
Then  subtracting  CAN  and  BAS  each  67°i  from  180,  we 
have  the  angle  "CAB=45°.  And  subtracting  CBN'  ll°i 
from  BAS  or  its  equal  ABN',  we  have  ABC=56°i.  The 
angle  at  C  is  therefore  78°  43'.     And 

Sin  C  :  AB;  :Sin  B  :  AC =7.63  miles. 
Example  IL 

New-York  iight-house  on  Sandy-Point  is  in  Lat.  40°  2S' 
N.  Lon.  74°  8'  W.  A  ship  observes  this  to  bear  N  76°  16' 
W.,  and  after  sailing  S  35°  10'  W.  8  miles,  finds  the  bearing 
to  be  N.  17°  13'  W.  Required  the  latitude  and  longitude  of 
the  ship,  at  ll^  first  observation. 

The  latitude  is         40°  26'J 
The  longitude  73    58} 

In  this  example,  as  the  difference  of  latitude  is  small,  the 
difference  of  longitude  is  best  calculated  by  middle  latitude 
sailing.  (Art.  74.) 

Example  III, 

A  merchant  ship  sails  from  a  certain  port  S.  51°  E.  at  the 
rate  of  8  miles  an  liour.  A  privateer  leaving  another  port  7 
mi'es  N  E  of  the  first,  sails  at  the  rate  ot  10  miles  an  hour. 
What  must  be  the  course  of  the  privateer,  to  meet  the  tihip. 
without  a  change  of  direction  in  either  ? 

Ans.  S.  7°  43'  E. 


CURRENT  SAILING.  49 

Example  IV, 

Two  light-houses  are  observed  from  a  ship  sailing  S.  38° 
W.  at  the  rate  of  6  miles  an  hour.  The  first  bears  N.  21° 
W.,  the  other  N.  47^  VV.  At  the  end  of  two  hours,  the  first 
is  found  to  bear  N.  5°  E.,  the  other  N.  13°  VV,  Wliat  is 
the  distance  of  the  hght-houses  from  each  other  f 

Ans.  G  miles  and  30  rods. 

IV.  Current  Sailing. 

87.  When  the  measure  given  by  the  log-line  is  taken  as 
the  rate  of  the  ship's  progress,  the  water  is  supposed  to  be 
at  rest.  But  if  there  is  a  tide  or  current,  the  log  being 
thrown  upon  the  water,  and  left  at  liberty,  will  move  with  it, 
in  the  same  direction,  and  with  the  same  velocity.  The  rate 
of  sailing,  as  measured  by  tlie  log,  is  the  motion  through  the 
water. 

If  the  ship  is  steered  in  the  direction  of  the  current,  her 
whole  motion  is  equal  to  the  rate  given  by  the  log,  added  to 
the  rate  of  the  current.  But  if  the  ship  is  iteered  in  opposi- 
tion to  the  current,  her  absolute  motion  is  equal  to  the  differ- 
ence between  the  current,  and  the  rate  given  by  the  log. 
In  all  other  cases,  the  current  will  not  only  affect  the  velocity 
of  the  ship,  but  will  change  its  direction. 

Suppose  that  a  river  runs  directly  south,  and  that  a  boat 
in  crossing  it  is  steered  before  the  wind,  from  west  to  east. 
It  will  be  carried  down  the  stream  as  fast,  as  if  it  were  mere- 
ly floating  on  the  water  in  a  calm.  And  it  will  reach  the  op- 
posite side  as  soon,  as  if  the  surface  of  the  river  were  at 
rest.     But  it  will  arrive  at  a  different  point  of  the  shore. 

Let  AB  (Fig.  26.)  be  the  direction  in  which  the  boat  is 
steered,  and  AD  the  distance  which  the  stream  runs,  while 
the  boat  is  crossing.  If  DC  be  parallel  to  AB,  and  BC  paral- 
lel to  AD ;  then  will  C  be  the  point  at  which  the  boat 
proceeding  from  A  will  strike  the  opposite  shore,  and  AC 
will  be  the  distance.  For  it  is  driven  across  by  the  wind,  to 
the  side  BC,  in  the  same  time  that  it  is  carried  down  by  the 
current,  to  the  line  DC 

In  the  same  manner,  if  Awt  be  zny  part  of  AB,  and  mn  be 
the  corresponding  progress  of  the  stream,  the  distance  sailed 
will  be  An.  And  if  the  velocity  of  the  ship  and  of  the  stream 
continue  uniforn),  Am  is  to  mn,  as  A B  to  BC^  so  that  AnC 

% 


60  NAVIGATION. 

is  a  straight  line,  (Euc   32.  6.)  The  lines  AB,  BC,  and  AC*. 
form  the  tliree  sidi  s  ol  a  triangle.     Hence, 

88.  If  the  direction  and  rate  of  a  ship's  motion  through 
the  water,  be  represented  by  the  position  and  length  of  one 
side  of  a  triangle,  and  the  direction  and  rale  of  the  current, 
by  a  second  sidr  ;  the  absolute  direction  and  distance  will  be 
shown  by  the  third  side. 

Example  I. 

If  the  breadth  of  a  river  running  south  (Fig.  28.)  be  300 
yards,  and  a  boat  steers  S.  75  E.  at  tht  rate  of  10  yards  in 
a  minute,  while  the  progress  of  the  stream  is  24  yards  in  a 
minute  ;  wl.at  is  the  actual  course,  and  what  distance  must 
the  boat  go  in  crossing  ? 

Cos  BaP  :  AP:  :R  :  AB=3io.6 
And      10  :  24: :  AB  :  BC  =  745.44 
Then  in  the  triangle  ABC, 
(BC+AB)  :  (BC-AB)::Tan   i   (BAC  +  BCA)  :  Tan  -I- 
(BAG  -  BCA)  =  i7°  33'  50" 

The  angle  BAC  is  55''  3'   50"     Then 
Sin  BAC  :  BC::Sin  ABC  :  AC'  =  879the  distance. 
AndDAC=BCA=19°  56   iO"  the  course. 

Example  IL 

A  boat  moving  through  the  water  at  the  rate  of  five  miles 
an  hour,  is  endeavouring  to  make  a  certain  point  lying  S, 
22i°  W.  while  the  tide  is  running  S.  78f°  E.  three  miles  an 
hour.  In  what  direction  must  the  boat  be  steered,  to  reach 
the  point  by  a  single  course .'' 

Ans.  S.  58°  33'  W. 

89.  But  the  most  simple  method  of  making  the  calculation 
for  the  effect  of  a  current,  in  common  cases,  especially  in  re- 
solving a  traverse,  is  to  consider  the  direction  and  rate  of  the 
current  as  an  additional  separate  course  and  distance;  and 
to  find  the  corresponding  departure  and  difference  of  latitude. 
A  boat  sailing  from  A  (Fig.  28.)  by  the  united  action  of 
the  wind  and  current,  will  arrive  at  the  same  point,  as  if  it 
were  first  carried  by  the  wind  alone  from  A  to  B,  and  then 
by  the  current  alone  from  B  to  C. 

Example  I. 

A  ship  sails  S.  17°  B.  for  2  hours,  at  the  rate  of  S  miles 


HADLEY'S  QUADRANT. 


51 


an  hour  ;  then  S.  18°  VV.  for  4  hours,  at  the  rate  of  7  miles 
an  hour  ;  and  during  the  whole  time,  a  current  sets  N,  76° 
W.  at  the  rate  of  2  miles  an  hour.  Required  the  direct 
course  and  distance. 


First  Course 
;Second  do. 
Current 

S.  17°  E. 
S.  18°  \V. 

N.  76°  W. 

Di,t, 

N. 

S. 

E. 

4.68 

W. 

16 
28 

1 2 

2.9 

\b.3 
26.6 

8.65 
11.64 
20.29 

4.68 

41.9 
2.9 

Dif.  Lat 

.39. 

D,p. 

15.61 

The  course  is  2*°  48'  50",  and  the  distance  42  miles. 
Example  IL 

A  ship  sails  S.  E.  at  the  rate  of  10  miles  an  hour  by  the 
lo^,  in  a  current  setting  E.  N.  E.  at  the  rate  of  5  miles  an 
hour.  What,  is  her  true  course  ?  and  what  will  be  her  dis- 
tance at  the  end  of  two  hours  ? 

The  course  is  66°  IJ',  and  the  distance  25.56  miles. 

V.    HaDLLY'S  QuADltANT. 

90.  In  the  preceding  sections,  has  been  particularly  ex- 
plained the  process  of  determining  the  place  of  a  ship  from 
her  course  and  distance,  as  given  by  the  compass  and  the 
log.  But  this  is  subject  to  so  many  sources  cf  errour,  from 
variable  winds,  irregular  currents,  lee-way,  uncertainty  of  the 
magnetic  needle,  &:c.  that  it  ought  not  to  be  depended  on, 
except  for  short  distances,  and  in  circumstances  which  forbid 
the  use  of  more  unerring  methods.  The  mariner  who  hopes 
to  cross  the  ocean  with  safety,  must  place  his  chief  reliance, 
for  a  knowledge  of  his  true  situation  from  time  to  time,  oa 
observatiofjs  of  the  heavenly  bodies.  By  these  the  latitude 
and  longitude  may  be  generally  ascertained,  with  a  sufficient 
degree  of  exactness.  It  belongs  to  astronomy  to  explain  the 
methods  of  making  the  calculations.  The  subject  will  not 
be  anticipated  in  this  place,  any  farther  than  to  give  a  de- 
scription of  the  quadrant  of  refl.xion.  commonly  called  Had- 
hifs  Quadrant ^^  by  which  the  altitudes  of  the  heaveniy  bo- 


*See  Note  H. 


b2  NAVIGATION. 

dies,  and  their  distances  from  each  other,  are  usually  measur-* 
ed  at  sea.  The  superiority  of  this,  over  most  other  astrono- 
mical instruments,  for  the  purposes  of  navigation,  is  owing  to 
the  fact,  that  the  observations  which  are  made  with  it,  are 
Jiot  materially  affected  by  the  motion  of  the  vessel, 

91.  In  explaining  the  construction  and  use  of  this  quadrant, 
it  will  be  necessary  to  take  for  granted  the  following  simple 
principles  of  Optics. 

1.  The  progress  of  light  when  it  is  not  obstructed,  or  turn- 
ed from  its  natural  course  by  the  influence  of  some  contigu- 
ous body,  is  in  n^A^  ^m€5.  Hence  a  minute  portion  of  light 
called  a  ray,  may  be  properly  represented  by  a  line. 

2.  Any  object  appears  in  the  direction  in  which  the  light 
from  that  object  strikes  the  eye.  If  the  light  is  not  made  to 
deviate  from  a  right  line,  the  object  appears  in  the  direction 
in  which  it  really  is.  But  if  the  light  is  reflected,  as  by  a 
common  mirror,  the  object  appears  not  in  its  true  situation, 
but  in  the  direction  of  the  glass,  from  which  the  light  comes 
to  the  eye. 

3.  TTie  angle  of  reflection  is  equal  to  the  angle  of  incidence; 
that  is,  the  angles  which  the  reflected  and  the  incident  rays 
make  with  the  surface  of  the  mirror,  are  equal  ;  as  are  also 
the  angles  which  they  make  with  a  perpendicular  to  the  mir- 
ror. 

92.  From  these  principles  is  derived  the  following  propo- 
sition ;  li^cn  light  is  reflected  by  two  mirrors  successively,  the 
angle  which  the  last  reflpcted  ray  makes  with  the  incident  ray, 
is  DOUBLE  the  angle  between  the  mirrors. 

If  C  and  D(Fig.  29.)   be   the  two   mirrors,  a  ray  of  light 
coming  from  A  to  C,  will  be  reflected  so  as  to  make  the   an- 
gle DCM  =  ACB  ;  and  will    be  again  reflected  at  D,  making 
HDM=CDE.     Continue  EC  and  ED  to  H,  draw  DG  pa- 
rallel to  BH,  and  continue  AC  to  P.     Then  is  CPM  the  an- 
gle which  the  last  reflected  ray  DP  makes  with  the  incident 
ray  AC  ;  and  DHM  is  the  angle  between  the  mirrors. 
By  the  preceding  article,  with  Euc.  29.  I  and  15.  1, 
GDC=DCM=ACB=PCM 
And  HDM=EDC=EDG-hGDC=DH]M-fPCM 
But  by  Euc.  32.  I  and  15.  1, 
rPM+PCM=DHM-fHDM=2DHM-{-PCM 
Therefore  CPM=2DHM 


HADLEY'S  QUADRANT.  53 

Cor.  1.  If  the  two  mirrors  make  an  angle  of  a  certain  num- 
ber of  degrees,  the  apparent  direction  of  the  object  will  be 
changed  twice  as  many  degrees.  The  object  at  A,  seen  by 
the  eye  at  P,  without  any  mirror,  would  appear  in  the  direc- 
tion PA.  But  after  reflection  from  the  two  mirrors,  the  light 
comes  to  the  eye  in  the  direction  DP,  and  the  apparent  place 
of  the  object  is  changed  from  A  to  R. 

Cor.  2.  If  the  two  mirrors  be  parallel^  they  will  make  no 
alteration  m  the  appirent  p'act?  of  the  object. 

93.  The  principal  parts  of  Hadley's  Quadrant  an;  the  fol- 
lowing ; 

1.  A  graduated  arc  kB  {V\g.  17.)  connected  with  the  ra- 
dii AC  and  BC. 

2.  An  index  CD,  one  end  of  which  is  fixed  at  the  centre  C, 
while  the  other  end  moves  over  the  graduated  arc. 

3.  A  plane  mirror  called  the  index  glass^  attached  to  the 
index  at  C.  Its  plane  passes  through  the  centre  of  motion 
C,  and  is  perpendicular  to  the  plane  of  the  instrument ;  that 
is,  to  the  plane  which  passes  through  the  graduated  arc,  and 
its  centre  C. 

4.  Two  other  plane  mirrors  at  E  and  M,  called  horizon 
glasses.  Each  of  these  is  also  perpendicular  to  the  plane  of 
the  instrument.  The  one  at  E,  called  the/ore  horizon  glass, 
is  placed  parallel  to  the  index  glass  when  the  index  is  at  0. 
The  other,  called  the  back  horizon  glass,  is  perpendicularto 
the  first  and  to  the  index  at  0.  This  is  only  used  occasional- 
ly, when  circumstances  render  it  difficult  to  take  a  good  ob- 
servation with  the  other. 

A  parf  of  each  of  these  glasses  is  covered  with  quicksilver, 
so  as  to  act  as  a  mirror  ;  while  another  part  is  left  transpar-' 
cnt,  through  which  objects  may  be  seen  in  their  true  situation' 

5.  Two  sight  vanes  at  G  and  L,  standing  perpendicular  to 
the  plane  of  the  instrument.  At  one  of  these,  the  eye  is 
placed  to  view  the  object,  by  looking  on  the  opposite  hori- 
zon glass.  In  the  fore  sight  vane  at  G,  there  are  two  perfo- 
rations, one  directly  opposite  the  transparent  part  of  the  fore 
horizon  glass,  the  other  opposite  the  silvered  part.  The  back 
sight  vane  at  L  has  only  one  perforation,  which  is  opposite 
the  centre  of  the  transparent  part  of  the  back  horizon  glass. 

6.  Coloured  glasses  to  prevent  the  eye  from  being  injurc^i 
by  the  dazzling  light  of  the  sun.     These  are  placed  at  H,  be- 


54  NAVIGATION. 

tween  the  index  mirror  and  the  fore  horizon  glass.  They 
may  be  taken  out  when  necessary,  and  placed  at  N  between 
the  index  mirror  and  the  back  horizon  glass. 

94.  This  instrument,  which  is  in  form  an  octant,  is  called  a 
quadrant,  because  the  graduation  extends  to  90  degrees,  al- 
though the  arc  on  which  these  degrees  are  marked  is  only 
the  eighth  part  of  a  circle.  The  light  coming  from  the  ob- 
ject is  first  reflected  by  the  index  glass  C,  (Fig.  J  7.)  and 
thrown  upon  the  horizon  glass  E,  by  which  it  is  reflected  to 
the  eye  at  G.  if  the  index  be  biought  to  0,  so  as  to  make 
the  index  glass  and  the  horizon  glass  parallel  ;  the  object 
will  appear  in  its  true  situation.  (Art.  92.  Cor.  2.)  But  if 
the  index  glass  be  turned,  so  as  to  make  with  the  horizon 
glass  an  angle  of  a  certain  number  of  degrees  ;  the  apparent 
direction  of  the  object  will  be  changed  twice  as  many  degrees* 

Now  the  graduation  is  adapted  to  the  apparent  change  in 
the  situation  of  the  object,  and  not  to  the  motion  of  the  in- 
dex. If  the  index  move  over  45  degrees,  it  will  alter  the  ap- 
parent place  of  the  object  90  degrees.  The  arc  is  common- 
ly graduated  a  short  distance  on  the  other  side  of  0  towards 
P.     This  partis  called  the  arc  of  excess, 

95.  The  quadrant  is  used  at  sea,  to  measure  the  angular 
distances  of  the  heavenly  bodies  from  each  other,  and  their 
elevations  above  the  horizon.  One  of  the  objects  is  seen  in 
its  true  situation,  by  looking  throuph  the  transparent  part  of 
the  horizon  glass.  The  other  is  seen  by  reflection,  by  look- 
ing on  the  silvered  part  of  the  same  glass.  By  turning  the 
index,  the  apparent  place  of  the  latler  may  be  changed,  till 
it  is  brought  in  contact  with  tiie  other.  The  motion  of  the 
index  whicliis  necessary  to  produce  this  change,  determines 
the  distance  of  the  two  objects.* 

9G.  To  find  the  distance  of  the  moon  from  a  star.  Hold  the 
quadrant  so  that  its  plane  shall  pass  through  the  two  objects. 
Look  at  the  star  through  the  transparent  part  of  the  horizon 
glass,  and  then  turn  the  index  till  the  nearest  edge  of  th  '  im- 
age of  the  moon  is  brought  in  contact  with  the  star.  This 
will  measure  the  distance  between  the  star  and  one  edcre  of 
tlie  moon.  By  adding  the  semi-diameter  of  the  moon,  we 
shall  have  the  distance  of  its  centre  from  the  star- 

*  For  the  a-f/Ui/z/icn/s  of  the  quadrant,  see  V'ince's  Practical  Astronomy, 
Maokay's  Navigation,  or  Bowditch's  Practical  Navigator. 


HADLEY'S  QUADRANT.  56 

The  distance  of  the  sun  from  the  moon,  or  the  distance  of 
two  stars  fronn  each  other,  nnay  be  measured  in  a  similar  man- 
ner. 

97.  To  measure  the  altitude  of  the  sun  above  the  horizon. 
Hold  the  instrument  so  that  its  plane  shall  pass  through  the 
sun.  ai  d  be  perpendicular  to  the  horizon  Then  move  the 
index  till  the  lower  edge  of  the  image  of  the  sun  is  brought 
in  contact  with  the  horizon,  as  seen  through  the  transparent 
part  of  the  glass. 

The  altitude  of  any  other  heavenly  body  may  be  taken  in 
the  same  manner. 

98.  To  measure  altitudes  by  the  back  observation.  When 
the  index  stands  at  o.  the  index  glass  is  at  right  angles  with 
the  back  horizon  glass.  (Art.  93.)  The  apparent  place  of  the 
object,  as  seen  by  reflection  from  this  glass,  must  therefore  be 
changed  ISO  degrees;  (Art.  92  Cor.  1.)  that  is.  it  must  ap- 
pear in  the  opposite  point  of  the  heavens.  In  taking  altitudes 
by  the  back  observation,  if  the  object  is  in  the  east,  the  ob- 
server faces  the  west  ;  or  if  it  be  in  the  south,  he  faces  the 
north  ;  and  moves  the  index,  till  the  image  formed  by  reflex- 
ion is  brought  down  to  the  horizon. 

This  method  is  resorted  to,  when  the  view  of  the  horizoR 
in  the  direction  of  the  object  is  obstructed  by  fog,  hills,  &ic. 

99.  Dip  or  Depression  of  the  Horizon.  In  taking  the  alti- 
tude of  a  heavenly  bodyat  sea,  with  Hadley's  Quadrant,  the 
reflected  image  of  the  object  is  made  to  coincide  with  the 
most  distant  visible  part  of  the  surface  of  the  ocean.  A  plane 
passing  through  the  eye  of  the  observer,  and  thus  touching 
the  ocean,  is  called  {\\q  marine  horizon  o{  i\\e  place  of  obser- 
vation If  BAB'  (Fig.  13.)  be  the  surface  of  the  ocean,  and 
the  observation  be  made  at  T,  the  marine  horizon  is  TA. 
But  this  is  different  from  the  true  horizon  at  T,  because  the 
eye  is  elevated  above  the  surface.  Considering  the  earth  as 
a  sphere,  of  which  C  is  the  centre,  the  true  horizon  is  TH 
perpendicular  to  TC  The  marine  horizon  TA  falls  below 
this.  The  angle  ATH  is  called  the  dip  or  depression  of  the 
horizon.  This  varies  with  the  height  of  the  eye  above  the 
surface.  Allowance  must  be  made  for  it,  in  observations  for 
determining  the  altitude  of  a  heavenly  body  above  ihe  true 
horizon. 

In  the  right  angled  triangle  ATC,  the  angle  ACT  is  equal 
to  the  angle  of  depression  ATH  ;  for  each  is  the  complement 


h6  NAVIGATION. 

of  ATC.  The  side  AC  is  the  semi-diameter  of  the  earth, 
and  the  hypothenuse  CT  is  equal  to  the  same  semi-diameter 
added  to  BT  the  height  of  the  eye.     Then 

AC  :  R:  :TC  :  Sec  ACT=ATH  the  depression.* 

100.  Artificial  Horizon.  Hadley's  Quadrant  is  particular- 
ly adapted  to  measurinji;  altitudes  at  sea.  But  it  may  be 
made  to  answer  the  same  purpose  on  land,  by  means  of  what 
is  called  an  artificial  horizon.  This  is  the  level  surface  of 
some  fluid  which  can  be  kept  perfectly  smooth.  Water  will 
answer,  if  it  can  be  protected  from  the  action  of  the  wind, 
by  a  covering  of  thin  glass  or  talc  which  will  not  sensibly 
change  the  direction  of  the  rays  of  light.  But  quicksilver, 
Barbadoes  tar,  or  clear  molasses,  will  not  be  so  liable  to  be 
disturbed  by  the  wind.  A  small  vessel  containing  one  of 
these  substances,  is  placed  in  such  a  situation  that  the  object 
whose  altitude  is  to  betaken  may  be  reflected  from  the  sur- 
hce.  As  this  surface  is  in  the  plane  of  the  horizon,  and  as 
the  angles  of  incidence  and  reflection  are  equal,  (Art.  91.) 
the  image  seen  in  the  fluid  must  appear  as  far  below  the  hori- 
zon, as  the  object  is  above.  The  distance  of  the  two  will, 
therefore,  be  double  the  altitude  of  the  latter.  This  distance 
may  be  measured  with  the  quadrant,  by  turning  the  index  so  as 
to  bring  the  image  formed  by  the  instrument  to  coincide  with 
that  formed  by  the  artificial  horizon. 

101.  77ie  Sextant  IS  SL  more  perfect  instrument  than  the 
quadrant,  though  constructed  upon  the  same  principle.  Its 
arc  is  the  sixth  part  of  a  circle,  and  is  graduated  to  120  de- 
grees. In  the  place  of  the  sight  vane,  there  is  a  small  tele- 
scope for  viewing  the  image.  There  is  also  a  magnifying 
glass,  for  reading  off* the  degrees  and  minutes.  It  is  common- 
ly made  with  more  exactness  than  the  quadrant,  and  is  bet- 
ter fitted  for  nice  observations,  particularly  for  determining 
longitude,  by  the  angular  distances  of  the  heavenly  bodies. 

A  still  more  accurate  instrument  for  the  purpose  is  the 
Circle  of  Reflexion.  For  a  description  of  this,  see  Borda  on 
the  Circle  of  Reflexion,  Rees'  Cyclopedia,  and  Bowditch's 
Bractical  Navigator. 

*  See  Note  I,  and  Table  II. 


SURVEYING. 


SECTION  I. 


SURVEYING  A  FIELD  BY  MEASURING  ROUND  IT. 

\rt.  105. T^^  "^^^^  common  method  of  surveying  a  fieid 

is  to  measure  the  length  of  each  of  the    sides, 

and  the  angles  which  they  make  with  the  meridian.     The 

lines  are  usually  measured  with  a  chain,  and  the  angles  with 

a  compass. 

106.  The  Compass,  The  essential  parts  of  a  Surveyor's 
Compass  are  a  graduated  circle,  a  magnetic  needle, and  sight- 
lioles  for  taking  the  direction  of  any  object.  There  are  fre- 
quently added  a  spirit  level,  a  small  telescope,  and  other  ap- 
pendages. The  instrument  is  called  a  Theodolite,  Circum- 
lercntor,  &zc.  according  to  the  particular  construction,  and 
the  uses  to  which  it  is  applied. 

For  measuring  the  angles  which  the  sides  of  a  field  make 
with  each  other,  a  graduated  circle  with  sights  would  be  suf- 
ficient. But  a  needle  is  commonly  used  for  determining  the 
position  of  the  several  lines  with  respect  to  the  meridian. 
This  is  important  in  running  boundaries,  drawing  deeds,  (fee 
It  is  true,  the  needle  does  not  often  point  directly  north  or 
south.  But  allowance  may  be  made  for  the  variation,  when 
this  has  been  determined  by  observation.     See  Sec.  V, 

107.  The  Chain,  The  Surveyor's  or  Gunter's  chain  is 
four  rods  long,  and  is  divided  into  100  li?iks.  Sometimes  a 
half  chain  is  used,  containing  50  links.  A  rod,  pole,  or  perch, 
is  161  feet.     Hence 

1    Link  =7.92  inches  =  |ofa  foot  nearly. 

1     Rod     =   25  links   =16-'- feet. 

1    Chain  =  100  links  =06  feet. 


58  SURVEYING. 

108.  The  measuring  unit  for  the^rra  of  a  field  Is  \hc  acre, 
which  coiitain?  160  square  rods.  If  then  (he  contents  iu 
square  rod?  be  divided  by  160,  the  quotient  will  he  (he  num- 
ber of  acres.  P/Ut  it  is  commonly  most  convenient  to  make 
the  computation  for  the  area  in  square  chains  or  links,  which 
are  decimals  of  an  acre.  For  a  square  chain  =4X4  =  16 
squnre  rods,  which  is  the  tenth  part  of  an  acre.  And  a  square 
link  =T^oXTio  =  Toiroe  of  »  ^q^^re  chain  =-j-^^u__  of  an 
acre.     Or  thus, 

625  links,  or  272{  feet  =  1  square  rod, 
10000  4356  =  1  chain  or  16  rods, 

250C0  10890  =  1  rood  or  40  rods, 

100000  43560  =  1  acre  or  160  rods. 

109.  The  contents,  then,  being  calculated  in  chains  and 
links  ;  if/oM/  places  of  decimals  be  cut  off,  the  remaining  fig- 
ures will  be  square  chains  ;  or  \i  five  places  be  cut  off,  the 
remaining  figures  will  be  acres.  Thus  the  square  of  16.32 
chains,  or  1632  links,  is  2663424  square  links,  or  266.3424 
square  chains,  or  26  63424  acres.  If  the  contents  be  con- 
sidered as  square  chains  and  decimals,  removing  the  decimal 
point  one  place  to  the  left  will  give  the  acres. 

1 10.  In  surveying  a  piece  of  land,  and  calculating  its  con- 
tents, it  is  necessary,  in  all  common  cases,  to  suppose  it  to 
be  reduced  to  a  horizontal  level.  If  a  hill,  or  any  uneven 
piece  of  ground,  is  bought  and  sold  ;  the  quantity  is  compu- 
ted, not  from  the  irregular  surface,  but  from  the  level  base 
on  which  the  whole  may  be  considered  as  resting.  In  run- 
ning the  lines,  therefore,  it  is  necessary  to  reduce  (hem  to  a 
level.  Unless  this  is  done,  a  correct  plan  of  the  survey  can 
never  be  exhibited  on  paper. 

If  a  line  be  measured  upon  an  ascent  which  is  a  regular 
plane,  though  oblique  to  the  horizon  ;  the  length  of  the  cor- 
responding level  base  may  be  found,  by  taking  the  angle  of 
elevation. 

Let  AB  (Fig.  30.)  be  parallel  to  the  horizon,  BC  perpen- 
dicular to  AB,  and  AC  a  line  measured  on  the  side  of  a  hill. 
Then,  the  angle  of  elevation  at  A  being  taken  with  a  quad- 
rant, (Art.  4.) 

R  :  Cos  A::AC  :  AB,  that  is, 
.^.'^  radius,  to  the  cosine  of  the  angle  of  elevation  ; 
So  is  the  oblique  line  measured}  to  the  corresponding  hori- 
.zontnl  base. 


SURVEYING.  59 

if  (he  chain,  instead  of  being  carried  parallel  to  the  surface 
of  the  ground,  be  kept  constantly  parallel  to  the  horizon  ;  the 
line  thus  measured  will  be  the  base  line  required.  The  line 
AB  (Fig.  50.)  is  evidently  equal  to  the  sum  of  the  parallel 
lines  abf  cd,  and  eC. 

Plotting  a  Survey. 

111.  When  the  sides  of  a  field  are  measured,  and  their 
bearings  taken,  it  is  easy  to  lay  down  a  plan  of  it  on  paper. 
A  north  and  south  line  is  drawn,  and  with  a  line  of  chords, 
a  protractor,  or  a  sector,  an  ancjle  is  laid  off,  equal  to  the  an- 
gle which  the  first  ?ide  of  the  field  makes  with  the  meridian, 
and  the  length  of  the  side  is  taken  from  a  scale  of  equal  oarts. 
(Trig.  15G — ir.l.)  Through  the  extremity  of  this,  a  serood 
meridian  is  drawn  parallel  to  the  first,  and  another  side  i;-  laid 
down  ;  from  the  end  of  this,  a  third  side,  &,c.  till  the  plan  is 
completed.  Or  the  plot  may  be  constructed  in  the  sane 
manner  as  a  traverse  in  navigation.  (Art.  7o.)  If  the  field  is 
correctly  surveyed  and  plotted,  it  is  evident  the  extremity  of 
the  last  side  must  coincide  with  the  beginnmg  of  the  first. 

Example  I. 


Draw  a  plan  ( 

of  a  field, 

from  the 

following 

courses 

and  dis 

fances, 

as  noted 

in  the  field-book  ; 

Ch. 

Links. 

1. 

N. 

7G° 

E. 

2 

46 

2. 

S. 

1G° 

W. 

3 

54 

3. 

N. 

8.3° 

W. 

2 

72 

4. 

N. 

12° 

E. 

2 

13 

5. 

N. 

60i°E. 

0 

95 

Let  A  (Fig.  31  )be  the  first  cornerof  the  field. 
Thro' A,  drawthemerid   N-S,  make  B  AN  =  78^  &:  \B  =  2.4G 
Thro'  B,draw  N'S' par.  to  NS,make  S'BC=16°,&;  BC  =  .S.54 
Thro'  C,drawN"S''par.  toNS,  makeDCN  =  83°,ii  CD=2.72 

&c.  &c. 

112.  To  avoid  the  inconvenience  of  drawing  parallel  lines, 
the  sides  of  a  field  may  be  laid  down  from  the  ans^les  v.hich 
they  make  ivith  each  other,  instead  of  the  angles  which  they 
make  with  the  meridian.  The  position  of  the  line  BC  (Fig. 
31.)  is  determined  by  the  angle  ABC,  as  well  as  by  the  an- 
gle S'BC.     When  the  several  courses  are  given,  the  angles 


60  :^LKV  EYING. 

which  any  two  conliguous  sides  make  with  each  other,  may 
be  known  by  the  following  rules. 

1.  Ifone  course  is  North  and  the  other  South,  one  East 
and  the  other  West  ;  subtract  the  less  from  the  greater, 

2.  Ifone  is  North  and  the  other  South,  but  both  East  or 
West ;  add  them  together, 

3.  If  both  are  North  or  South,  but  one  East  and  the  other 
West  ^subtract  their  sum  from  ISO  degrees, 

4.  If  both  are  North  or  South,  and  both  East  or  West  ; 
add  together  90  degrees,  the  less  course^  and  the  complement  of 
the  greater. 

The  reason  of  these  rules  will  be  evident  by  applying  thenn 
to  the  preceding  example.  (Fi^.  31.) 

The  first  course  is  BAN,  which  is  equal  to  ABS'.  (Euc. 
29.1.)  If  from  this  the  second  course  CBS'  be  subtracted, 
there  will  remain  the  an^jle  ABC. 

If  the  second  course  CBS',  or  its  equal  BCN",  be  added 
to  the  third  course  DCN  ;  the  sum  will  be  the  angle  BCD. 

The  sum  of  the  angles  CDS,  NDE,  and  CDE,  is  180  de- 
grces.  (Euc.  13.1.)  If  then  the  two  first  be  subtracted  from 
180  degrees,  the  remainder  will  be  the  angle  CDE. 

Lastly,  let  EP  be  perpendicular  to  NS.  Then  the  sum  of 
the  angles  DES,  PES,  and  AEP  the  complement  of  AEN, 
is  equal  to  the  angle  DEA. 

We  have  then  the  angle  ABC=G2°,  DEA  =  131i°, 
BCD=&9°,  EAB  =  162i°. 
CDE=85°, 

With  these  angles,  the  field  may  be  plotted  without  draw- 
ing parallels,  as  in  Trig.  173. 

Finding  the  Contents  of  a  Field. 

113.  There  are  in  common  use  two  methods  of  finding  the 
contents  of  a  piece  of  land,  one  by  dividing  the  plot  into  tri- 
angles, the  other  by  calculating  the  departure  and  dijfer^nce 
of  latitude  for  each  of  the  sides. 

When  a  survey  is  plotted,  the  whole  figure  may  be  divided 
into  triangles,  by  drawing  diagonals  from  the  different  an- 
gles. The  lengths  of  the  diagonals,  and  of  the  perpendicu- 
lars on  the  bases  of  the  triangles,  may  be  measured  on  the 
same  scale  of  equal  parts  from  which  the  sides  of  the  field 
were  laid  down.     The  area  of  each  of  the  triangles  is  equal 


SURVEYING.  6.1 

(o  half  the  product  of  its  base   and  perpendicular;  and  their 
sum  is  the  area  of  the  whole  figure.  (Mens.  13.) 
Example  I, 

Let  the  plan  Fig.  32  be  the  same  as  Fig.   31,  the  sides  of 
which,  with  their  bearing?,  are  given  in  art.  111. 

Then  the  triangle  ABC  =  BCx^AP  =3.84  sq.  chains. 
ACE  =  ACxiEP'=1.53 
DCE  =  ECxiDP"=2.89 

The  contents  of  the  whole  =  8.26 
1 1 4.  This  method  cannot  be  relied  on,  where  great  accur- 
acy is  required,  if  the  lines  are  measured  by  a  scale  and  com- 
passes only.  But  the  parts  of  the  several  triangles  may  be 
found  by  trigonometrical  calculation^  independently  of  the 
projectiorj  ;  and  then  the  area  of  each  may  be  computed, 
either  from  two  sides  and  the  included  angle,  or  from  the 
three  sides.  (Mens.  9,  10.) 

The  sides  of  the  iield  and  their  bearings  being  given  by  the 
survey,  the  angles  of  the  original  (igure  may  all  be  known. 
(Art.  112.)  Tiicn  in  the  triangle  ABC  (Fig.  32.)  we  have  the 
sides  AB  and  BC,  with  the  angle  ABC,  to  find  the  other 
parts.  (Trig.  153.)  And  in  the  triangle  CDE,  we  have  the 
sides  DC  and  DE,  with  the  angle  CDE.  Subtracting  the 
angle  BAC  from  BAE,  we  shall  have  CAE  ;  and  subtracting 
DEC  from  DEA,  we  shall  have  CEA.  There  will  then  be 
given,  in  the  triangle  ACE,  the  side  EA  and  the  angles. 
(Trig.  150.) 

The  sides  and  bearings,  as  given  in  art.  Ill,  are 

1.  AB      N.  70°  E.  2.4G  chains. 

2.  BC      S.   16    VV.3.54 

3.  CD     N    03    W.  2.72 

4.  DE     N.   12   E.  2.13 

5.  EA      N.  GO}yE,  0.95 
Then  by  Mensuration,  art.  9, 

R  :  Sin  ABC::ABXBC  :  2  are«  ABC  =  7.69  sq.  chains 
R  :  Sin  AEC::AExEC  :  2  area  AEC=3.0G 
R  :  Sin  CDE:  :CDxDE  :  2  area  CDE =5.77 


2)16.52 


Contents  of  the  whole  field,  8.26 


62  SURVEYING. 

Or  the  areas  of  tlie  several  triant^les  may  be  found  by  the 
rule  in  Mensuration,  art.  10,  viz.  If  a,  b,  and  c,  be  the  sides 
of  any  triangle,  and  /t=ha!f  their  sum  ; 

The  area  =\/hX{h  —  u)X{h- b)X{h  —  c) 

Example  11. 

Courses.  Ch.      Links. 

1.  E.  2G      34 

2.  S.  10°  30' E.   32      26 

3.  N.42         W.   13       35 

4.  S.  58  W.   23      52 

5.  N.  30      55 
Contents  of  the  field,  69.735  acres. 

The  method  which  has  been  explained,  of  ascertaining  the 
contents  of  a  piece  of  land  by  dividing  it  into  triangles,  is  of 
use  in  cases  which  do  not  require  a  greater  degree  of  accura- 
cy, than  can  be  obtained  by  the  scale  and  compasses.  But 
if  the  areas  of  the  triangles  are  to  be  found  by  trigonometri- 
cal calculation,  the  process  becomes  too  laborious  for  com- 
mon practice.  The  following  method  is  often  to  be  preferred. 

Finding  the  area  of  a  field  by   departure   and  differ- 
ence OF  latitude. 

115.  Let  ABCDE  (Fig.  33.)  be  the  boundary  of  a  field. 
At  a  given  distance  from  A,  draw  the  meridian  line  NS. 
Parallel  to  this  draw  L'R',  AG,  BH,  and  DK.  These  may 
be  considered  as  portions  of  meridians  passing  through  the 
points  A,  B,  D,  and  E.  For  all  the  meridians  which  cross  a 
field  of  moderate  dimensions,  may  be  supposed  to  be  paral- 
lel, without  sensible  errour.  At  right  angles  to  ISSdraw  the 
parallels  AL,  BM,  CO,  EP,and  DR.  These  will  divide  the 
figure  LABCDR  into  the  three  trapezoids  ABML,  BCOM, 
and  CDRO  ;  and  the  figure  LAEDR  into  the  two  trapezoids 
DEPR  and  EALP.  The  area  of  the  field  is  evidently  equal 
to  the  difference  between  these  two  figures. 

The  sum  of  the  parallel  sides  of  a  trapezoid,  multiplied  in- 
to their  distance,  is  equal  to  twice  the  area.  (Mens.  12.) 
Thus 

(AL-\-BM)x  AG =2  area  ABML. 

Now  AL  is  a  given  distance,  and  BM=AL-f  BG.  But 
BG  is  the  departure^  and  AQ  the  difference  of  latitude,  cor- 


SURVEYING.  63 

responding  to  AB  one  of  the  sides  of  the  field.  (Arts.  39,  40.) 

And  bj  art  44, 

r>    1     T^-  J.    A  t> .  .  ^  Sin  BAG  :  Depart.  BG 
Rad  :  D.st.  AB. .  |  ^^^  ^^q  .  jj.^  Lat.  AG 

Or  the  departure  and  difference  of  latitude  may  be  taken 
from  the  Traverse  Table,  as  in  Navigation.  (Art.  50.) 

In  the  same  manner,  from  the  sides  BC,  CD,  DE,  and  EA, 
may  be  found  the  deparlnre  CH,  CK.  DR',  AL'.  and  the  dif- 
ferences of  latitude  BH,  DK,  ER',  and  EL'.  We  shall  then 
have  the  parallel  sides  of  each  of  the  trapezoids,  or  the  dis- 
tances of  the  several  corners  of  the  field  from  the  meridian 
NS.     For 

BM  =  AL4-BG,  DR=CO-CK, 

CO=BM+CH,  EP=DR-DR'. 

If  the  field  be  measured  in  the  direction  ABCDE,  the  dif- 
ferences of  latitude  AG,  BH,  and  DK,  will  be  SouthingSj 
v/hile  R'E  and  E'L  will  be  Korihings,  The  former  are  the 
breadths  of  the  three  trapezoids  which  form  the  figure 
LABCDR  ;  and  the  latter  are  the  breadths  of  the  two  trape- 
zoids which  form  the  figure  LAEDR  The  difference,  then 
between  the  sum  of  the  products  of  the  northings  into  the  cor- 
responding meridian  distances,  and  the  sum  of  the  products 
of  the  southings  into  the  corresponding  meridian  distances, 
is  twice  the  area  of  the  field. 

It  will  very  much   facilitate  the  calculation,  to  place  in  a 
fable  the  several  courses,  distances,  northings,  southings,  &c 
We  have,  then,  the  following 

Rule. 

IIG.  Find  the  northing  or  southing,  and  the  easting  or  west- 
i.ng,  for  each  side  of  the  field,  and  place  them  in  distinct  col" 
urnns  in  a  table.  To  these  add  a  column  of  Meridian  Distan- 
ces,for  the  distance  of  one  end  of  each  side  of  the  field  from  a 
given  meridian  ;  a  column  of  Multipliers  to  contain  the  pairs 
of  meridian  distances  for  the  two  ends  of  each  of  the  sides  ^  and 
columns  for  the  north  and  south  Areas,  See  Fig.  23,  and  the 
table  for  example  1. 

Suppose  a  meridian  line  to  be  drawn  without  the  field,  at  any 
given  distance  from  the  first  station  ;  and  place  the  assumed 
distance  at  the  head  of  the  column  of  Meridian  Distances,  To 
this  add  the  first  departure^  if  both  be  east  or  both  io.cst  ;  but 
,'^ifbtract,  if  one  be  east  and  the  other  west  ;  and  place  the  sum 


64  SUKVEYING. 

or  difference  in  ike  column  of  Meridian  Distances^  against  the 
first  course.  To  or  from  the  last  numhcr^  add  or  subtract  the 
second  departure^  ^c.  4'c. 

For  the  column  of  Multipliers^  add  together  the  first  and  5f- 
cond  numbers  in  the  column  of  Meridian  Distances  ;  the  se- 
cond and  thirds  the  third  and  fourth  j  ^c.  placing  the  sums  op- 
posite the  several  courses. 

Mullipli^  each  7iumber  in  the  column  of  Multipliers  into  its 
corresponding  northing  or  southing,  aiid place  the  product  in  the 
column  oj  north  or  south  areas.  The  difference  between  the 
sum  of  the  north  areas,  and  the  sum  of  the  south  areas,  zoill  he 
twice  the  area  of  the  field. 

This  method  of  finding  the  contents  of  a  field,  as  it  depends 
on  departure  and  difference  of  latitude,  which  are  calculated 
by  right-angled  trigonometry,  is  sometimes  called  Rectangular 
Surveying. 

117.  If  the  assumed  meridian  pass  through  the  eastern  or 
western  extiemity  of  the  field,  as  L'ER'  (Fig.  33.)  the  dis- 
tance EP  will  be  reduced  to  nothing,  and  the  figures  AEL' 
and  EDR'  will  be  triangles  instead  of  trapezoids.  If  the  sur- 
vey be  made  to  begin  at  the  point  E,  cipher  is  to  be  placed 
at  the  head  of  the  column  of  meridian  distances,  and  the  first 
number  in  the  column  of  multiplier?  will  be  the  same,  as  the 
first  in  the  column  of  meridian  distances.     See  example  II. 

1 18.  When  there  is  a  re-entering  angle  in  a  field,  situated 
with  respect  to  the  meridian  as  CDE  ;  (Fig.  34.)  the  area 
EDM,  being  included  in  the  tigure  BCRA,  will  be  repeated  in 
the  column  of  south  areas.  But,  as  it  is  also  included  in  the 
figure  DCRM,  it  will  be  contained  in  the  column  of  north 
areas.  Therefore  the  difference  between  the  north  areas  and 
the  south  areas,  will  be  twice  the  area  of  the  field,  in  this 
case,  as  well  as  in  others. 

119.  If  any  side  is  directly  east  or  loest,  there  will  be  no 
difference  of  latitude,  and  consequently  no  number  to  be 
placed  against  this  course,  in  the  columns  of  north  and  south 
areas.     See  example  11.  Course  1.  AB  (Fig.  34.) 

The  number  in  the  columns  of  areas  will  be  wanting  also, 
when  any  side  of  the  field  coincides  with  the  assumed  roe 
rrdian.     See  example  II.  Conne  5.  EA  (Fig.  34.) 


SURVEYING.  65 

120.  In  finding  the  departure  and  difference  of  latitude 
from  the  traverse  table,  the  numbers  for  the  links  may  be 
looked  out  separately;  care  bcin^  taken  to  remove  the  deci- 
mal point  two  places  to  the  left,  because  a  link  is  the  100th 
part  of  a  chain. 

Thus  if  the  course  be  29°,  and  the  distance  23.46  chains  ; 

The  dif.  of  lat.  &;depart.  for  23  chains  are  20. 1 2  and  11.15 
for  46  links  .40  .22 


for  23.46 


20.52 


11.3: 


Example  I,     See  Fig.  33. 


Covrus. 

Ditt. 

Dif.  Lot 

Departure. 

M  D. 

A     L 
.0    E. 

Mult 

l^  ^tat 

o      .            1 

N.     i     S. 

E    :  w. 

it.     BAG 

S.  64°  E. 

AB 

30  cA. 

I     AG 

1    13  15 

,     GB 

26. 9f. 

BM 

46.96 

ALtBM 

66.96 

.Ai«.UL 
880  5346 

2.    CBH 

S.  14°  K 

BC 
10 

1     BH 
i     9.70 

HC 

2.42 

CO 

493-< 

BM+CO 

96  34 

2B.M0C 

934  4980 

3.    CDK 
S.  35°  W 

CD 
30 

KD 

24.57 

1 

CK 

17  21 

DR 

82.(7 

COfDR 
8i  55 

;;CDRO 

2003  •:83i( 

4.    KDE 

N.  65°  W. 

DE 

R'E  1 
8.45    1 

DR' 

18.13 

EP 

14.04 

DR+EP 

2  DEPR 
390.4715 

¥ 

5.    L'EA 

N.  8«42'E 

EA 
39.42 

EL'  ; 
38.97  : 

LA 

5. 96 

AL 

JO 

EP+AL 
34.04 

1  2EPLA 
1 1326  5388 

i7^2 

47  42 

35.34 

35JJ 

1 

J|717.ll3:i 

3818.7055 

Twice  the  figure  ABCDRL  is  3818.7055  square  chains  ; 
Twice  the  figure  AEDRL        1717.1133 

The  difference  2101.5922 

The  contents  of  the  field      1050.7961  sq.  ch.  or  105.0796 

acres.     (Art.  109.) 


10 


o6 


SURVEViNG. 
Example  IL    See  Fig.  S-i, 


Couries. 

/)«/. 

Dijf.  Lai. 

N.    1    S. 

Depaiiurt. 
E.     1    VV. 

M. 
Dist. 

00 

Mult. 

N.Areas. 

S.  Areas 

1.    E. 

AC 

26.34 

00 

13.G4 

30.55 
44,19 

1 
00 

31.72 

12.47 
44.19 

26.34 

5.8r 

00 

1 

12.27 

19.95 

00 

AB 

26.34 

CR 

32.22 

DM 

19.95 

00 
00 

00 

00 

00 

2.S.10rE. 

BC 

32.26 

AB+CR 

58.56 

CR+PM 
52.17 

2ABCR 

1857.5232 

3.  N.42^VV. 

CD 

18.3.3 

2CD.\^R 

711.598» 

4.S.58"\V. 

DE 

23.52 

DM 

19.95 

2DME 

248.7765 

5    i\. 

EA 
30.55 

00 

1 

1 

711.5988 

2106.29971 

The  contents  of  the  field  =^(2106.3-71 1.6)=:697..35  sq.  ch. 

Or  69.735  acres. 
In  this  example,  the  meridian  distance  of  the  first  station 
A  being  nothing,  cipher  is  placed  at  the  head  of  the  column 
of  meridian  distances.  (Art.  117.)  The  first  side  AB  being 
directly  east  and  west,  has  no  difference  of  latitude,  and 
therefore  the  numbcrin  the  column  of  areas  against  this  course 
is  wanting,  as  it  is  against  the  fifth  course,  which  is  directly 
north.  (Art.  119.)  The  number  against  the  fourth  course, 
in  the  column  of  multipliers,  is  only  the  length  of  the  line 
DM;  the  figure  DME  being  a  triangle^  instead  of  a  trape- 
zoid. 

Example  III, 

Find  the  contents  of  a  field  bounded  by  the  following 


lines; 


1.  N.  SS''   30'  E.  15  ch.  50Hnks. 


2. 

N.  72 

45  E.   18 

70 

3. 

S.  70 

45  E.   18 

70 

4. 

S.  53 

W.  12 

45 

5. 

S.  83 

15  E.  24 

10 

6. 

S.  31 

15  W.15 

20 

7. 

S.  62 

45  W.  22 

60 

8. 

N.  73 

30  VV.27 

30 

9. 

N.  17 

25  W.  14 

50 

The  area  is  145^  acres. 


S.URVEYING.  ,37 

\2l.  When  a  field  is  correctly  surveyed,  and  the  depar- 
tures and  differences  of  latitude  accurately  calculated;  it  ij^ 
evident  the  sum  of  the  northings  must  be  equal  to  the  sum  of 
the  southings,  and  the  sum  of  the  eastings  equal  to  the  sum 
of  the  westings.  Jf  upon  adding  up  the  numbers  in  the  de- 
parture and  latitude  columns,  the  northings  are  not  found  to 
agree  nearly  with  the  southings,  ;.nd  the  eastings  with  the 
westings,  there  must  be  an  crrour,  either  in  the  survey  or  in 
the  calculation,  which  requires  that  one  or  both  should  be 
revised.  But  if  the  difference  be  small,  and  if  there  be  no 
particular  reason  for  supposing  it  to  be  occasioned  by  one 
part  of  the  survey  rather  than  another  '  it  may  be  apportion- 
ed among  the  several  departures  or  differences  of  latitude, 
according  to  the  different  lengths  of  the  sides  of  the  tield,  by 
the  following  rule  ; 

As  the  whole  perimeter  of  the  field, 
To  the  whole  errour  in  departure  or  latitude  ; 
So  is  the  length  of  one  of  the  sides, 
To  the  correction  in  the  corresponding  departure  or 
latitude, 

This  correction,  if  applied  to  the  column  in  whieh  the  sum 
of  the  numbers  is  too  small,  is  to  be  added ^  but  if  to  the  oth> 
er  column,  it  is  to  be  subtracted,*  See  the  example  on  the 
next  page. 


*Seo  the  fourth  Number  of  the  \x)H]yst  publi?hpd  of  Phflaflelphin. 


■—^•1- 

—i 

B 
0 

•n    - 

i 

-vj   Oi   Or   ^   03   ♦O   — 

tz;  fy3  ^  CA^  c/D  c/3  !2; 

•0  to  03  ►t*'  0  to  C;t 

0       0*.|«*|-      0,0- *.i- 

00           00 

j^  5=1  <  ?i  <  w  p 

Courses. 

; 

1^  CO  to  !^  ^  lu  ^ 

Ml-  b»|-                       M|- 

2 

2. 

b 

OS 

c 

-t 

03 

Ci 

0 

to                             — 

p           03                          p 

CO         Cr                     to 

GO 

Oi  OJ 

Oi  CD 

0   *t^ 

6.70 

8.43 

13.96 

5.85 

-1 

o 

c 

to 

CO 

0 

to 

CO 

0 

•—        to  ^ 

b       CO  ^ 

►;^        -v)  CO 

to 

■i 

p 

«o 

JO  ;;-J  to         -vl 

ot  iii.  b       b 

OS  CO  0         -J 

i 

1 

i 

0 
CO 

I 

-h  1 1 1  1  j  t 

b  b  b  b  b  b  b 

•4   00    »f^    Cji    fc^    Cjt    05 

rp 

r*  r' 

1    1    1  4-1  ff 

c>  '0  '0  h  (D  '0  c> 

CO  fi^  o-  cr.  >f^  --■*  00 

bo 

0:1 

to                               — 

p          03                         p 
Id         in                       03 

? 

^ 

b 

CO       b  03  b 

to          —  CO  Cj' 

K3 
CO 

Id 

^           CO   CO 

0              J:^     -v>      , 

1 

CO 

CO 

7.03 

11.95 

7.45 
2.48 

KJ   ^0   to   — 

Q  JO   CO   --   p   ^   rfi- 

0   I^-   CO   CO    <>   CO   CO 

CO   0.    CO   CO   H-   ^ 

0 
0 

»-*  oo  Oi'  ku  t^i.  — 

to   to   1—  »0  CO  fO  ^^ 

ii.   4:^.   CO   b  6^    b   CO 
CO   —    —    C5   CO   CO   -J 

1 

♦* 

Oj 

b 
to 

153.46 

112.61 
51.85 

r^ 

C/0 

b 

00 

o» 
to 

Cjt   4:1.   to 
^        CO  0  CO 

JO           03   ^    00 

to       ^  b  c> 

03           0   -J   to 

6* 

SURVEYING.  ^9 

In  this  example,  the  whole  perimeter  of  the  field  is  IOO4. 
chains,  the  whole  errour  in  latitude  .34,  the  whole  errour  in 
departure  .42.  and  the  length  of  the  first  side  18.  To  find 
;he  corresponding  errours, 

lOOi  'IS'  *  \'*^^  '  '^^  ^^^®  errour  in  latitude, 
2  *       ' '  1 .42  :  .08  the  errour  in  departure. 

The  errour  in  latitude  is  to  be  added  to  10.26  making:  it 
10.32,  as  in  the  column  of  corrected  northings;  and  the  er- 
rour in  departura  is  to  be  added  to  14.79  making  it  14.87,  as 
in  the  column  of  corrected  eastings.  After  the  corrections 
are  made  for  each  of  the  courses,  the  remaining  part  of  the 
calculation  is  the  same  as  in  the  precedmg  examples. 

122.  If  the  length  and  direction  of  each  of  the  sides  of  a 
field  except  one  be  given,  the  remaining  side  may  be  easily 
found  by  calculation.  For  the  difference  between  the  sum 
of  the  northings  and  the  sum  of  the  southing?  of  the  ?iven 
sides,  is  evidently  equal  to  the  northing  or  southing  of  the  re- 
maining side ;  and  the  difference  between  the  sum  of  the 
eastings  and  the  sum  of  the  westings  of  the  given  sides,  is 
equal  to  the  easting  or  westing  of  the  remaining  side.  Hav- 
ing then  the  difference  of  latitude  and  departure  lor  the  side 
required,  its  length  and  direction  may  be  found,  in  the  same 
manner  as  in  the  sixth  case  of  plane  sailing.     (Art.  49.) 

Example  V, 

What  is  the  area  of  a  field  of  six  sides,  of  which  five  are 
<^iven,  viz. 


1. 

S. 

56° 

E. 

4,18  chains 

2. 

N. 

21 

E. 

480 

3. 

N. 

56 

W. 

3.06 

4. 

S. 

21 

W. 

0.13 

5. 

6. 

Th( 

N. 
I  ai 

66«i 

W. 

1.44 

ea  is 

two 

acres. 

Exampl 

e  F7. 

1.  N    38°  W.  17.21   chains 

2.  N.  13  E.  21.16 

3.  N    72   E.  24.11 

4.  S.   41    E.  19  26 

5.  S.   11    W.  2435 

6.  


70  SURVEYIN«. 

123.  Plotting  hy  departure  and  difference  of  Latitude.  A 
survey  may  be  easily  plotted  from  the  northings  and  south- 
ings, eastings  and  westings.  For  this  purpose,  the  column  of 
Meridian  Distances  is  used.  It  will  be  convenient  to  add  al- 
so another  column,  containing  the  distance  of  each  station 
from  a  given  parallel  of  latitude,  and  /ormed  by  adding  the 
northings  and  subtracting  the  southings,  or  adding  the  south- 
ings and  subtracting  the  northings. 

Let  AT  (Fig.  33.)  be  a  parallel  of  latitude  passing  through 
the  first  station  of  the  field.  Then  the  southing  TB  or  LM  is 
the  distance  of  B,  the  second  station,  from  the  given  parallel. 
To  this  adding  the  southing  BH,  we  have  LO  the  distance 
of  CO  from  LT.  Proceeding  in  this  manner  for  each 
of  the  sides  of  the  field,  and  copying  the  7th  column  in  the 
table,  p.  65,  we  have  the  following  differences  of  latitude  and 
meridian  distances. 

Diff.  Lot,  Merid,  BisU 

AL  20 

1.  LM  13.15  BM  46.9G 

2.  LO  22.85  CO  49  38 

3.  LR  47  42  DR  32.17 

4.  LP  38.97  EP   14.04 

To  plot  the  field,  draw  the  meridian  NS,  and  perpendicu- 
lar to  this,  ibe  parallel  of  latitude  LT.  From  L  set  off  the 
differences  of  latitude  LM,  LO,  LR,  and  LP.  Through 
L,  M,  O,  R,  and  P,  draw  lines  parallel  to  LT;  and  set  off 
the  meridian  distances  AL,  BM,  CO,  DR,  and  EP.  The 
points  A,  B,  C,  D,  and  E,  will  then  be  given. 

124.  When  a  field  is  a  regular  figure,  cs  a  parallelogram, 
triangle,  circle,  &ic.  the  contents  may  be  found  by  the  rules 
in  Mensuration,  Sec.  L  and  IL 

125.  The  area  of  a  field  which  has  been  plotted,  is  some- 
times found  by  reducing  the  whole  to  a  TiiiANfiLB  of  the  same 
area.  This  is  done  by  changing  the  figure  in  such  a  manner 
as,  at  each  step,  to  make  the  number  of  sides  one  less,  till  they 
are  reduced  to  three. 

Let  the  side  AB  (Fig.  35.)  be  extended  indefinitely  both 
ways.  To  reduce  the  two  sides  BC  and  CD  to  one,  draw  a 
line  from  D  to  B,  and  another  parallel  to  this  from  C,  to  in- 
tersect AB  continued.  Draw  also  a  line  from  D  to  the  point 
of  intersection  G.     Then  the  triangles  DBC  and  DBG  are 


SURVEYING.  7i 

equal.  (Euc.  37.  1.)  Taking  from  cacli  the  common  part 
DBH,  there  remains  BGH  equal  to  DCH.  If  then  the 
triangle  DCH  be  thrown  out  of  the  plot,  and  BGH  be  added, 
we  shall  have  the  five-sided  figure  AGDEF  equal  to  the  six- 
sided  figure  ABCDEF. 

In  the  same  manner,  the  line  EL  may  be  substituted 
for  the  two  sides  AF  and  EF ;  and  then  DM,  for  EL  and 
ED  This  will  reduce  the  whole  to  the  triangle  MGD, 
which  is  equal  to  the  original  figure.  The  area  of  the  trian- 
gle may  then  be  found  by  multiplying  its  base  into  half  its 
height;  and  this  will  be  the  contents  of  the  field. 

In  practice  it  will  not  be  necessary  actually  to  draw  the 
parallel  lines  BD,  GC,  he.  It  will  be  sufficient  to  lay  the 
edge  of  a  rule  on  C,  so  as  to  be  parallel  to  a  line  supposed  to 
pass  through  B  and  D,  and  to  mark  the  point  of  intersec- 
tion G. 

126.  If  after  a  field  has  been  surreyed,  and  the  area  com- 
puted, the  chain  is  found  to  be  too  long  or  too  short ;  the 
true  contents  may  be  found,  upon  the  principle  that  similar 
figures  are  to  each  other  as  the  squares  of  their  homologous 
sides.  (Euc.  20.6.)  The  proportion  may  be  stated  thus  ; 

As  the  square  of  the  true  chain,  to  the  square  of  that  by 

which  the  survey  was  made  ; 
So  is  the  computed  area  of  the  field,  to  the  true  area. 

Ex  If  the  area  of  a  field  measured  by  a  chain  66.4  feet 
Jong,  be  computed  to  be  32.6036  acres ;  what  is  the  area  as 
measured  by  the  true  chain  66  leet  long.'* 

Ans.  33  acres. 

127.  A  plot  of  a  field  may  be  changed  to  a  different  scale, 
that  is,  it  may  be  enlarged  or  diminished  in  any  given  ratio, 
by  drawing  lines  parallel  to  each  of  the  sides  of  the  original 
plan. 

To  enlarge  the  perimeter  of  the  figure  ABCDE  (Fig.  36.) 
in  the  ratio  of  aG  to  AG ;  draw  lines  from  G  through  each  of 
the  angular  points.  Then  beginning  at  n,  drawai  parallel  to 
AB,  &c  parallel  to  BC,  &c. 


72  SURVEYING. 

It  is  evident  that  the  Gtigles  are  the  same  in  the  enlarged 
figure,  as  in  the  original  one.     And  by  similar  triangles, 

AG  :aG::BG  :  6G::CG  :  cG::&c. 

And 

AG  :  aG::AB  :  ab::BC  :bc::hc. 

Therefore  ABCDE  and  abode  are  similar  figures.  (Euc. 
Def   1    6.) 

In  the  same  manner,  the  smaller  figure  a'b'dd'e'  may  be 
drawn,  so  as  to  have  its  perimeter  proportioned  to  ABCDE 
as  a'G  to  AG. 


SECTION  n. 

Methods  of  Surveying  in  particular  cases, 


A  i^ft  T^MEASURING  round  a  field,  in  the  manner 
IT JL  explained  in  the  preceding  section,  is  by 
far  the  most  common  method  of  surveying.  The  following 
problems  are  sometimes  useful.  They  may  serve  to  verify 
or  correct  the  surveys  which  are  made  by  the  usual  method. 

Problem   I. 
To  survey  a  field  from  two  stations. 
129.  Find  the  distance  of  the  two   stations,  and 

THEIR  bearings  FROM  EACH  OTHER  ;  THEN  TAKE  THE  BEAR 
INGS  OF  THE  SEVERAL  CORNERS  OF  THE  FIELD  FROM  EACH 
OF  THE  STATIONS. 

In  the  field  ABCDE,  (Fig.  37.)  let  the  distance  of  the  two 
Stations  S  and  T  be  given,  and  their  bearings  from  each  other. 
By  taking  the  bearing  of  A  from  S  and  T,  or  the  angles 
AST  and  ATS,  we  have  the  direction  of  the  lines  drawn 
from  the  two  stations  to  one  of  the  corners  of  the  field. 
The  point  A  is  determined  by  the  intersection  of  these  lines. 
In  the  same  manner,  the  point  B  is  determined,  by  the  inter- 
section of  SB  and  TB;  the  point  C,  by  the  intersection  of 
SC  and  TC  ;  &c.  &tc.  The  sides  of  the  field  are  then  laid 
down,  by  connecting  the  points  ABCD,  &£c. 

The  area  is  obtained,  by  finding  the  areas  of  the  several 
triangles  into  which  the  field  is  divided  by  lines  drawn  from 
one  of  the  stations.  Thus  tlie  area  of  AbCDE  (Fig.  37.)  is 
equal  to 

ABT-fBCT+CDT+DET-fEAT 

or  to 
ABS+BCS-fCDS  +  DES+EAS 

Now  wc  have  the  base  line  ST  given  and  the  angles,  in  the 
triangle  AST,  to  find  AS  and  AT  j  in  the  triangle  BST,  to 
find  BS  and  BT,  &ic.     After  these  are  found,  we  have  two 

n 


74  .^L'KVKVINCI. 

sides  and  the  included  angle  in  the  triangles  ABT,  BCT,  &:o. 
from  which  the  areas  may  be  calculated.  (Mens.  9.) 

Example. 

Let  the  station  T  (Fig.  37.)  be  N.  80^  E.  from  S,  the  dis- 
tance ST  27  chains,  and  the  bearings  of  the  several  corners 
of  the  field  from  S  and  T  as  follows ; 

TA  N.  30°  W.  SA  N.  17°  E. 

TB  N.  15    E.  SB  N    55   E. 

TC  S.  53    E.  SC  S.   73   E. 

TD  S.  65    W.  SD  S.    24    W. 

TE  N.  70    W.  SE  N.  26    W. 

Tiiese  will  give  the  following  angles  j 
ATS=  70°  AST=  63°  ATB=  46* 

BTS  =  115  BST=  25  BTC  =  U2 

CTS  =  133  CST=  27  CTD=I08 

DTS=  25  DST  =  124  DTE=  55 

ETS=:  30  EST =106  ETA=  40 

From  which,  with  the  base  line  ST,  are  calculated  the  fol- 
lowing lines  and  areas. 

AT=32.89  chains.  ABT  =  206.45  sq.  chains 

BT=  17.75  BCT  =  294.95 

CT  =  35.84  CDT  =  740.7 

DT=43.46  DET=665  1 

ET  =  37.3G  EAT  =  395. 


Contents  of  the  field,       =230.22  acres. 

The  course  and  length  of  each  of  the  sides  of  the  field  may 
be  found,  if  necessary.  After  the  parts  mentioned  above  are 
calculated,  there  will  be  given  two  sides  and  the  included  an- 
gle, in  the  triangle  ATB,  to  find  AB,  in  BTC  to  find  BC 
he. 

If  the  base  line  between  the  two  stations  be  too  short,  com- 
pared with  the  sides  of  the  field  and  their  distances,  the  sur- 
vey will  be  liable  to  inaccuracy.  It  should  not  generally  be 
less  than  one  tenth  of  the  longest  straight  line  which  can  be 
drawn  on  the  ground  to  be  measured. 

130.  It  is  not  necessary  that  the  base  line,  from  the  extre- 
mities of  which  the  bearings  are  taken,  should  be  within  the 
field.  It  may  be  one  of  the  sides,  or  it  may  bo  entirely  with- 
out the  field. 


SURVEYING.  75 

Let  S  and  T  fFig.  38.)  be  two  stations  from  wiilch  all  the 
corners  of  a  field  ABODE  may  be  seen.  If  the  direction 
and  length  of  the  base  line  be  measured,  and  the  bearings  of 
the  points  A,  B,  C,  D,  and  E,  be  taken  at  each  of  the  sta- 
tions;  the  areas  of  the  several  triangles  may  be  found.  The 
6gure  ABCTDE  is  equal  to 

DET+EAT+ABT+BCT 

From  this  subtracting  DCT,  we  have  the  area  of  the  field 
ABCDE. 

In  this  manner,  a  piece  of  ground  may  be  measured  which, 
from  natural  or  artificial  obstructions,  is  inaccessible.  Thus 
an  island  may  be  measured  from  the  opposite  bank,  or  an 
enemy's  camp,  from  a  neighbouring  eminence. 

131.  The  method  of  surveying  by  making  observations 
from  two  stations,  is  particularly  adapted  to  the  measurement 
of  a  bay  or  harbour. 

The  survey  may  be  made  on  the  water,  by  anchoring  two 
vessels  at  a  distance  from  each  other,  and  observing  from 
each  the  bearings  of  the  several  remarkable  objects  near  the 
shore.  Or  the  observations  may  be  made  from  such  eleva- 
ted situations  on  the  land  as  are  favourable  for  viewing  the 
figure  of  the  harbour.  If  all  the  parts  of  the  shore  cannot  be 
seen  from  two  stations,  three  or  more  may  be  taken.  In  this 
case,  the  direction  and  distance  of  each  from  one  of  the  oth- 
ers should  be  measured. 


Proble 


:*i. 


To  survey  a  field  by  measuring  from  one  station. 
132.  Take  the  bearings  of  the  several  corners  of 

THE  field,  and  MEASIJRE  THE  DISTANCE  OF  EACH   FROM  THE 
GIVEN   STATION. 

If  the  length  and  direction  of  the  several  lines  AT,  BT, 
CT,  DT,  and  ET,  (Fig.  37.)  be  ascertnined  ;  there  will  be 
given  two  sides  and  the  included  angle  of  each  of  the  trian- 
gles ABT,  BCT,  CDT,  DET,  and  EAT  ;  from  which  their 
areas  may  be  calculated,  (Mens.  9.)  and  the  sum  of  thesQ 
will  be  the  contents  of  the  whole  figure. 


76  SURVEYINC:;. 

The  station  may  be  taken  in  one  of  the  sides  or  angles  of 
the  field,  as  at  C.  (Fig.  32.)  The  lines 

CD,  CE,  CA,  CB,  and  the  angles 
DCE,  EGA,  ACB,  being  given, 
the  areas  of  the  triangles  nnay  be  found.  ^ 


Problem   III. 
To  survey  a  field  by  the  chain  alone. 

133.  Measure  the  SIDES  of  the  field,  and  the  DI- 
AGONALS BY  WHICH  IT  IS  DIVIDED  INTO  TRIANGLES. 

By  measuring  the  sides  (Fig.  32.) 

AB,  BC,  CD,  DE,  EA, 

and  the  diagonals  CA  and  CE,  we  have  the  three  sides  of 
each  of  the  triangles  into  which  the  whole  figure  is  divided. 
They  may  therefore  be  constructed,  (Trig.  172.)  and  theii 
areas  calculated.  (Mens.  lO.) 

134.  To  measure  an  angle  with  the  chain,  set  off  equal 
distances  on  the  two  lines  which  include  the  angle,  as  AB, 
AC,  (Fig.  39.)  and  measure  the  distance  from  B  to  C.  There 
will  then  be  given  the  three  sides  of  the  isosceles  triangle 
ABC,  to  firrd  the  angle  at  A  by  construction  or  calculation. 

The  chain  may  be  thus  substituted  for  the  compass,  in  sur- 
veying a  field  by  going  round  it,  according  to  the  method 
explained  in  the  preceding  section  ;  or  by  measuring  from 
one  or  two  stations,  as  in  problems  I.  and  II, 

Problem  IV. 

To  survey  an  irregular  boundary  by  means  of  offsets. 

135.  Run  a  straight  line  in  ant  convenient  direc- 
tion, AND  MEASURE  THE  PERPENDICULAR  DISTANCE  OF  EACH 
ANGULAR  point  OF  THE  BOUNDARY  FROM  THIS  LINE. 

The  irregular  field  (Fig.  40.)  may  be  surveyed,  by  taking 
the  bearing  and  length  of  each  of  the  four  lines  AE,  EF, 
FI,  lA,  and  measuring  the  perpendicular  distances  BB', 
CC,  DD',  GGr',  HH',  KK'.  These  perpendiculars  are  cal- 
led offsets,     h  is  necessary  to  note  in  a  field  book  the  parts 


PURVEYING 


77 


into  which  the  line  that  is  measured  is  divided  by  the  offsets^ 
as  in  the  following  example.     (See  Fig.  40.) 


Offsets  on 

the  left        1       Courses  ctnd  Distances.        \ 

Offsetts  on  the  right. 

ChaiDB. 

AEN.85°E.  12.74  ch. 

BB' 

2.18 

AB'     3.25 

CC 

2.18 

BC    2.13 

DI> 

1.23 

CD'    1.12 
DE    6.24 

1 

EF  S.  24°  E.  7.23 

Fl  N.  87  W.  13.34 

GG' 

2.66 

FG     3.84 

HH' 

1.48 

G'H'  2.22 
H'l     7.28 

— ■ 

lA  N.  26  VV.  5.32 

IK' 

KK'  2.94 

K'A 

As  the  offsets  are  perpendicular  to  the  lines  surveyed,  the 
little  spaces  ABB',  BB'CC,  CC'DD'  he.  are  either  right  an- 
gled triangles,  parallelograms,  or  trapezoids.  To  find  the 
contents  of  the  field,  calculate  in  the  first  place  the  area  be- 
tween the  lines  surveyed,  as  the  trapezium  AEFIA,  (Fig.  40.) 
and  then  add  the  spaces  between  the  offsets,  if  they  fall  with- 
in the  boundary  line ;  or  subtract  them,  if  they  fall  without, 
as  AIK. 

When  any  part  of  a  side  of  a  field  is  inaccessible,  equal  off- 
sets may  be  made  at  each  end,  and  a  line  run  parallel  to  the 
boundary. 

Problem  V, 


To  measure  the  distance  between  any  two  points  oh  the  sur- 
face of  the  earth,  by  means  of  a  series  of  triangles  extending 
from  one  to  the  other. 

136.  Measure  a  side  of  one  of  the  triangles  for  a 
BASE  LINE,  TAKE  THE  BEARING  of  this  or  some  oth- 

HR    SIDE,    AND    MEASURE    THE    ANGLES    IN    EACH  OF    THE 
TRIANGLES. 

If  it  be  required  to  find  the  distance  between  the  two 
points  A  and  I,  (Fig.  41.)  so  situated  that  the  measure  can- 
not be  taken  in  a  direct  line  from  one  to  the  other;  let  a  se- 


78  SURVEYING. 

ries  of  triangles  be  arranged  in  such  a  manner  between  them, 
that  one  side  shall  be  common  to  the  first  and  second,  as 
BC,  to  the  second  and  ihird  as  CD,  to  the  third  and  fourth, 
&tc.  Then  measure  the  length  of  BC  for  a  base  line,  take 
the  bearing  oi"  the  side  AB,  and  measure  the  angles  of  each 
of  the  triangles. 

These  data  are  sufficient  to  determine  the  length  and  bear- 
ing of  each  of  the  sides,  and  the  distance  and  bearing  of  I 
from  A.  For  in  the  two  first  triangles  ABC  and  BCD.  the 
angles  are  given  and  the  side  BC,  to  find  the  other  sides. 
When  CD  is  found,  there  are  given,  in  the  third  triangle 
CDE.  one  side  and  the  angles,  to  find  the  other  side.  In 
the  same  manner,  the  calculation  may  be  carried  from  one 
triangle  to  anoiher,  till  all  the  sides  are  found. 

The  bpiirifigs  of  the  sides,  that  is,  the  angles  which  they 
make  with  the  meridian,  may  be  determined  from  the  bear- 
ing of  the  first  side,  and  the  angles  in  the  several  triangles. 
Thus  if  NS  be  parallel  to  AM,  the  angle  BAP,  or  its  equal 
ABN  subtracted  from  ABD  leaves  NBD;  and  this  taken 
from  180  degrees  leaves  SBD. 

From  the  bearing  and  length  of  AB  may  be  found  the 
southing  AP,  and  the  easliiigr  PB.  In  the  same  manner  are 
found  the  several  southings  PP,  P'P",  P"P"',  P"'M.  The 
sum  of  the  southings  is  the  line  AM.  And  if  the  distance  is 
so  smfill,  that  the  several  meridians  may  be  considered  paral- 
lel, the  difference  between  the  sum  of  the  eastings  and  the 
sum  of  the  westings,  is  the  perpendicular  IM.  We  have 
then,  in  the  right-angled  triangle  AMI,  the  sides  AM  and  MI, 
to  find  the  distance  and  bearing  of  I  from  A. 

137.  This  problem  is  introduced  here,  for  the  purpose  of 
giving  the  general  outlines  of  those  important  operations 
which  have  been  carried  on  of  late  years,  with  such  admira- 
ble precision,  under  the  name  of  Trigonometrical  Surveying* 

Any  explanation  of  the  subject,  however,  which  can  be 
made  in  this  part  of  the  course,  must  bo  very  imperfect.  In 
the  demonstration  of  the  problem,  the  several  triangles  are 
supposed  to  be  in  the  same  plane,  and  the  distances  of  the 
meridians  so  small,  that  they  may  be  considered  parallel.  But 
in  practice,  the  ground  upon  which  the  measurement  is  to  be 
made  is  very  irregular.  The  stations  selected  for  the  angular 
points  of  the  triangles,  are  such  elevated  parts  of  the  country 
as  are  visible  to   a   considerable    distance.     They    should 


SURVEYING.  7;. 

be  so  situated,  that  a  signal  staff,  tower,  or  other  conspicuous 
object  in  any  one  of  the  angles,  may  be  seen  from  tl)e  other 
two  angles  in  the  same  triangle.  It  will  rarely  be  the  case 
that  any  two  of  the  triangles*  will  be  in  thi>  same  plane,  or 
any  one  of  them  parallel  to  the  horizon.  Reductions  will 
therefore  be  necessary  to  bring  them  to  a  common  level. 
But  even  this  level  is  not  a  plane  In  the  cases  in  which  this 
kind  of  surveying  is  commonly  practised,  the  n]easnrement 
is  carried  over  an  extent  of  country  of  many  miles.  The  se- 
veral points,  when  reduced  to  the  same  distance  from  the 
centre  of  the  earth,  are  to  be  considered  as  belonging  to  a 
spherical  surface.  To  make  the  calculations  then,  if  the  line 
to  be  measured  is  of  any  considerable  extent,  and  if  nice  ex- 
actness is  required,  a  knowledge  of  Spherical  Trigonometry 
is  necessary. 

13S.  The  decided  superiority  of  this  method  of  surveying, 
in  point  of  accuracy,  over  all  others  which  have  hitherto  been 
tried,  particularly  where  the  extent  of  ground  is  great,  is  ow- 
ing partly  to  the  fact  that  almost  all  the  quantities  measure<i 
are  angles;  and  partly  to  this,  that  for  the  single  line  which 
it  is  necessary  to  measure,  the  ground  maif  he  chosen,  any 
where  in  the  vicinity  of  the  system  of  triangles.  It  would  be 
next  to  impossible  fo  determine  the  precise  horizontal  dis- 
tance between  two  points,  by  carrying  a  chain  over  an  irregu- 
lar surface.  But  in  the  trigonometrical  measurements 
which  arc  made  upon  a  great  scale,  there  can  generally  be 
found,  somewhere  in  the  country  surveyed,  a  level  plane,  a 
heath  or  a  body  of  ice  on  a  river  or  lake,  of  sufficient  extent 
ibr  a  base.  This  is  the  only  line  which  it  is  absolutely  neces- 
sary to  measure.  It  is  usual  however,  to  measure  a  second, 
which  is  called  a  line  of  verification.  If  the  length  of  the 
base  BC  (Fig.  41.)  and  the  angles  be  given,  all  the  other 
lines  in  the  figure  may  be  found  by  trigonometrical  calcula- 
tion. But  if  GH  be  also  measured,  it  will  serve  to  detect 
any  errour  which  may  have  been  committed,  cither  in  taking 
ihe  angles,  or  in  computing  the  sides,  of  the  series  of  trian-. 
gles  between  BC  and  GH, 

139.  In  measuring  these  hues,  rods  of  copper  or  platina 
have  been  u>cd  in  France,  and  glass  tubes  or  steel  chains  in 
England.  The  results  have  in  many  instances  been  extreme- 
ly exact.  A  base  was  measured,  on  Hounslow  Heath,  by 
General  Roy,  with  glass  rods.     Several  years  after,  it  was  re- 


80  SURVEYING. 

measured  by  Colonel  Mudge,  wilh  a  steel  chain  of  very  nice 
construciion.  The  dilFerence  in  the  two  measurements  was 
5css  than  three  inches  in  more  than  five  miles.  Two  parties 
measured  a  base  in  Peru  ot  62T5i  toises,  or  more  than  seven 
miles;  and  the  difference  in  their  results  did  not  exceed  two 
inches. 

Exact  as  these  measurements  are,  the  exquisite  construc- 
tion of  the  instruments  which  have  been  used  for  taking  the 
angles,  has  given  to  that  part  of  the  process  a  still  higher  de- 
gree of  perfection.  The  amount  of  the  errours  in  the  angles 
of  each  of  the  triangles,  measured  by  Ramsden's  The- 
odolite, did  not  exceed  three  seconds.  In  the  great  surveys  in 
France,  the  angles  were  taicen  with  nearly  the  same  cor- 
rectness. 

140.  One  of  the  most  important  applications  of  trigono- 
metrical surveying,  is  in  meat-uring  ores  of  the  meridian,  or  of 
parallels  of  latitude,  particularly  the  former.  This  is  neces- 
sary in  determining  itie  figure  of  the  earth,  a  very  essential 
problem  in  Geography  and  Astronomy.  A  degree  of  zeal 
has  been  dispiu)ed  on  this  subject,  proportioned  to  its  prac- 
tical importance.  Arcs  of  the  meridian  have  been  measured 
at  great  expense,  in  England,  France,  Lapland,  Peru,  &ic. 
Men  of  distinguished  science  have  engaged  in  the  under- 
taking. 

A  meridian  line  has  been  measured,  under  the  direction  of 
General  Roy  and  Colonel  Mudge,  from  the  Isle  of  Wight,  to 
Clifton  in  the  north  of  England,  a  distance  of  about  200 
miles.  Several  years  were  occupied  in  this  survey.  An- 
other arc  passing  near  Paris,  has  been  carried  quite  through 
France,  and  even  across  a  part  of  Spain  to  Barcelona.  In 
measuring  this,  several  distinguished  mathematicians  and  as- 
tronomers were  engaged  for  a  number  of  years.  These  two 
arcs  have  been  connected  by  a  system  of  triangles  running 
across  the  English  Channel,  the  particular  object  of  which 
was  to  determine  the  exact  difference  of  longitude  between 
the  observatories  of  Greenwich  and  Paris.  Besides  the  me- 
ridian arcs,  other  lines  intersecting  them  in  various  directions 
have  been  measured,  both  in  England  and  France.  With 
these,  the  most  remarkable  objects  over  the  face  of  the 
country  have  been  so  connected,  that  the  geography  of  the 
various  parts  of  the  two  kingdoms  is  settled,  wilh  a  precision 
which  could  not  be  expected  from  any  other  method. 


SURVEYING.  81 

141.  The  exactness  of  the  surveys  will  be  seen  from  a 
comparison  of  the  lines  of  verification  as  actually  measured, 
with  the  lengths  of  the  same  lines  as  determined  by  calcula- 
tion. These  would  be  affected  by  the  amount  of  all  the  er- 
rours  in  measuring  the  base  lines,  in  taking  the  angles,  in 
computing  the  sides  of  the  triangles,  and  in  making  the  ne- 
cessary reductions  for  the  irregularities  of  surface.  A  base 
of  verification  measured  on  Romney  Marsh  in  England,  was 
found  to  differ  but  about  two  feet  from  the  length  of  the  same 
line,  as  deduced  from  a  series  of  triangles  extending  more 
than  60  miles.  A  base  of  verification  connected  with  the 
meridian  passing  through  France,  was  found  not  to  differ  one 
foot  from  the  result  of  a  calculation  which  depended  on  the 
measurement  of  a  base  400  miles  distant  A  line  of  verifi- 
cation of  more  than  7  miles,  on  Salisbury  Plain,  differed 
scarcely  an  inch  from  the  length,  as  computed  from  a  system 
of  triangles  exiending  to  a  base  on  Hounslow  Heath.* 


♦  S««  Note  K. 


IS 


SECTION  111. 

LAYING  OUT  AND  DIVIDING  LANDS. 


A   '      142    nr^  ^^^°^^  ^^^°  ^^^  familiar  with  the  principles  of 
^  '  '  geometry,  it  will  be  unnecessary  to  give  par- 

ticular rules,  for  all  the  various  methods  of  dividing  and  lay- 
ing out  lands.  The  following  problems  may  serve  as  a  spe- 
cimen of  the  manner  in  which  the  business  may  be  conduct- 
ed in  practice. 

Problem  I. 

To  lay  out  a  given  number  of  acres  m  the  form  of  a  square. 

143.  Reduce  the  number  of  acres  to  s(^uare  rods  ob 

CHAINS,  and  EXTllACT  THE  SQ,UARE  ROOT.    This  will  givC  OnC 

side  of  the  required  field.     (Mens.  7.) 

Ex.  1.  What  is  the  side  of  a  square  piece  of  land  con- 
taining 1241  acres.  Ans.  141  rods. 

2.  What  is  the  side  of  a  square  field  which  contains  58f 
acres  ? 

Problem  IL 

To  lay  out  a  field  in  the  form  of  a  parallelogram,  when  one 
side  and  the  contents  are  given, 

144.  Divide  THE  number  of  square  rods  or  chains  by 
THK  LE^GTH  OF  THE  GIVEN  SIDE.  The  quotient  will  be  a 
side  perpendicular  to  the  given  side.     (Mens.  7.) 

Ex.  What  is  the  width  of  a  piece  of  land  which  is  286 
rods  long,  and  which  contains  77  acres  ? 

Ans.  44  rods. 

Cor.  As  a  triangle  is  half  a  parallelogram  of  the  same  base 
and  heiglit,  a  field  may  be  laid  out  in  the  form  of  a  triangle 
whose  area  and  base  are  given,  by  dividing  twice  the  area  by 
the  base.  The  quotient  will  be  the  perpendicular  from  the 
opposite  angle.     (Mens.  8.J 


SURVEYING.  83 


Problem  III. 

To  lay  out  a  p^iece  of  land  in  the  form  of  a  parallelograjn,  the 
length  of  which  shall  be  to  the  breadth  in  a  given  ratio. 

145,  As  the  length  of  the  parallelogram  to  its  breadth  ; 
So  is  the  area,  to  the  area  of  a  square  of  the  same 

breadth. 

The  side  of  the  square  may  then  be  found  by  problem  I, 
and  the  length  of  the  parallelogram  by  problem  II. 

If  BCNM  (Fig.  42.)  is  a  square  in  the  right  parallelogram 
ABCD,  or  in  the  obh'que  parallelogram  ABC'D',  it  is  evi- 
dent that  AB  is  to  MB  or  its  equal  BC,  as  the  area  of  the 
parallelogram  to  that  of  the  square. 

Ex.  If  the  length  of  a  parallelogram  is  to  its  breadth  as 
7  to  3,  and  the  contents  are  52^  acres,  what  is  the  length  and 
breadth. 

Problem  IV. 

TTie  area  of  a  parallelogram  being  given,  to  lay  it  out  in  such 
a  form,  that  the  length  shall  exceed  the  breadth  by   a  given 

DIFFERENCE. 

146.  Let  x=BC  the  breadth  of  the  parallelogram  ABCD 

(Fig.  42.)  and  the  side  of  the  square  BCNM. 
J=AM  the  difference  between  the  length  and 

breadth. 
a=the  area  of  the  parallelogram. 
Then  a  =  (x-{-d)Xx=x^-i'dx.     (xMens.  4.) 
Reducing  this  equation,  we  have 

\^a^fd^-  '^d=x. 
That  is,  to  the  area  of  the  parallelogram,  add  one  fourth 
of  the  square  of  the  difference  between  the  length  and  the 
breadth,  and  from  the  square  root  of  the  sum,  subtract  half 
the  difference  of  the  sides  ;  the  remainder  will  be  the  breadth 
of  the  parallelogram. 

Ex.  If  four  acres  of  land  be  laid  out  in  the  form  of  a 
parallelogram,  the  difference  of  whose  sides  is  12  rods,  what 
is  the  breadth  ? 


84  SURVEYING. 


Problem  V, 


To  lay  out  a  triangle  whose  area  and  angles  are  given, 

147.  Calculate  the  area  of  any  supposed   triangle 

WHICH  has  the  same  ANGLES.       THEN 

As  THE  AREA  OF  THE  ASSUMED  TRIATJGLE, 
To  THE  AREA  OF  THAT  WHICH   IS  REQUIRED  ; 
So   IS  THE  SQUARE  OF  ANY  SIDE  OF  THE  FORMER, 
To  THE   SQUARE  OF    THE    CORRESPONDING    SIDE     OF    THE 
LATTER. 

If  the  triangles  B'CC  and  BCA  (Fig.  43.)  have  equal  an- 
gles, they  are  similar  figures,  and  therefore  their  areas  are  as 
he  squanjs  of  their  like  sides,  for  instance,  as  AC  *  CC  • 
(Euc.  19.  6.)  The  square  of  CC  being  found,  extracting  the 
Square  root  will  give  the  line  itself 

To  lay  out  a  triangle  of  which  one  side  and  the  area  are 
given,  divide  twice  the  area  by  the  given  side  ;  the  quotient 
will  be  the  length  of  a  perpendicular  on  this  side  from  the 
opposite  angle.  (Mens.  8.)  Thus  twice  the  area  of  ABC 
(Fig.  45.)  divided  by  the  side  AB,  gives  the  length  of  the 
perpendicular  CP. 

148.  This  problem  furnishes  the  means  of  cutting  off,  or 
laying  out,  a  given  quantity  of  land  in  various  forms. 

Thus-  from  the  triangle  ABC,  (Fig.  43.)  a  smaller  triangle 
of  a  given  area  may  be  cut  off,  by  a  line  parallel  to  AB. 
The  line  CC  being  found  by  the  problem,  the  point  C  will 
be  given,  from  which  the  parallel  line  is  to  be  drawn. 

1 49.  If  the  directions  of  the  lines  AE  and  BD,  (Fig.  44.) 
and  the  length  and  direction  of  AB  be  given  ;  and  if  it  be 
required  to  lay  off  a  given  area,  by  a  line  parallel  to  AB  ;  let 
the  lines  AE  and  RD  be  continued  to  C.  The  angles  of  the 
triangle  ABC  with  the  side  AB  being  given,  the  area  may  be 
found.  From  this  subtracting  the  area  of  the  given  trapezoid, 
the  remainder  will  be  the  area  of  the  triangle  DCE  ;  from 
which  may  be  found,  as  before,  the  point  E  through  which 
the  parallel  is  to  be  drawn. 

If  the  trapezoid  is  to  be  laid  off  on  the  other  side  of  AB, 
its  area  must  be  added  to  ABC,  lo  give  the  triangle  D'CE'. 

150.  If  a  piece  of  land  is  to  be  laid  off  from  AB,  (Fig. 
45.  by  a  line  in  a  given  direction  as  DE,  7iot  parallel  to  AB  ; 


SURVEYING.  85 

let  AC  parallel  to  DE  be  drawn  through  one  end  of  AB 
The  required  trapezium  consists  of  two  parts,  the  triangf 
ABC,  and  the  trapezoid  ACED.  As  the  angles  and  one 
side  of  the  former  are  given,  its  area  may  be  found.  Sub- 
stracting  this  from  the  given  area,  we  have  the  area  of  the 
trapezoid,  from  which  the  distance  AD  may  be  found  by  the 
preceding  article. 

151.  If  a  given  area  is  to  be  laid  off  from  AB,  (Fig.  46.) 
by  a  line  proceeding  from  a  s^ivrn  point  D  ;  first  lay  off  the 
trapezoid  ABCD.  If  this  be  too  small,  add  the  triangle 
DCE  ;  but  if  the  trapezoid  be  too  large,  subtract  the  triangle 
DCE'. 

Problem  VI. 

To  divide  the  area  of  a  triangle  into  parts  having  given  ratios 
to  each  other,  by  lines  drawn  from  one  of  the  angles  to  the 
opposite  base. 

152.  Divide  the  base  in  the  same  proportion  as  the 

PARTS  required. 

If  the  triangle  ABC  (Fig.  47.)  be  divided  by  the  Hues  CH 
and  CD  ;  the  small  triangles,  having  the  same  height,  are  to 
each  other  as  the  bases  BH,  DD,  and  AD.     (Euc.  1.6,) 

Problem  VII. 

To  divide  an  irregular  piece  of  land  into  any  two  given  parts. 

153.  Run  a  line  at  a  venture^  near  to  the  true  division  line 
required,  and  find  the  area  of  one  of  the  parts.  If  this  be  too 
large  or  too  small,  add  or  subtract,  by  the  preceding  articles,  a 
triangle,  a  trapezoid,  or  a  trapezium,  as  the  case  may  require. 

A  field  may  sometimes  be  conveniently  divided  by  redu- 
cing it  to  a  triangle,  as  in  Art.  125,  (Fig.  35.)  and  then  divi- 
ding the  triangle  by  problem  VI. 


SECTION  IV. 

LEVELLLVG. 


A  134  \^  *^  frequently  necessary  to  ascertain  how 
much  one  spot  of  ground  is  higher  than  anoth- 
er. The  practicabihty  of  supplying  a  town  with  water  from 
a  neighbouring  fountain,  will  depend  on  the  comparative  ele- 
vation of  the  two  places  above  a  common  level.  The  di- 
rection of  the  current  in  a  canal  will  be  determined  by  the 
height  of  the  several  parts  with  respect  to  each  other. 

The  art  of  levelling  has  a  primary  reference  to  the  level 
surface  of  water.  The  surface  of  the  ocean,  a  lake,  or  a 
river,  is  said  to  be  level  when  it  is  at  rest.  If  the  fluid  parts 
of  the  earth  were  perfectly  ^pAenca/,  every  point  in  a  level 
surface  would  be  at  the  same  distance  from  the  centre.  The 
difference  in  the  heights  of  two  places  above  the  ocean  would 
be  the  same,  as  the  difference  in  their  distances  from  the 
centre  of  the  earth.  It  is  well  known  that  the  earth,  though 
nearly  spherical,  is  not  perfectly  so.  It  is  not  necessary, 
however,  that  the  difference  between  its  true  figure  and  that 
of  a  sphere  should  be  brought  into  account,  in  the  compara- 
tively small  distances  to  which  the  art  of  levelling  is  com- 
monly applied.  But  it  is  important  to  distinguish  between 
the  true  and  the  apparent  level. 

155.  The.  TRUE  LEVEL  is  a  CURVE  which  either  coincides 
Tcith,  or  is  parallel  to,  the  surface  of  water  at  rest. 

The  APPARENT  LEVEL  is  a  STRAIGHT  LINE  which  is  fl  TAN- 
GENT to  the  true  level j  at  the  point  where  the  observation  is 
made, 

Thusif  ED  (Fig.  48.)  be  the  surface  of  the  ocean,  and 
AB  a  concentric  curve,  B  is  on  a  true  level  with  A.  But  if 
AT  be  a  tangent  to  AB,  at  the  point  A,  the  apparent  level,  as 
observed  at  A,  passes  through  T. 

156.  When  levelling  instruments  are  used,  the  level  is  de- 
termined either  by  a/mrf,  or  a  plumb-line.  The  surface  of 
the  former  h  parallel  to  the  horizon.     The  latter  h  perpen- 


LEVELLING.  87 

dicular.  One  of  the  most  convenient  instrunoents  for  the 
purpose  is  the  spirit  level,  A  glass  tube  is  nearly  filled  with 
spirit,  a  small  space  being  left  for  a  bubble  of  air.  The  tube 
is  so  formed,  that  when  it  is  horizontal,  the  air  bubble  will  be 
in  the  middle  between  the  two  ends.  To  the  glass  is  at- 
tached an  index  with  sight  vanes  ;  and  sometimes  a  small 
telescope,  for  viewing  a  distant  object  distinctly.  The  sur- 
veyor should  also  be  provided  with  a  pair  of  levelling  rods, 
which  are  to  be  set  up  perpendicularly,  at  convenient  dis- 
tances, for  the  purpose  of  measuring  the  height  from  the  sur- 
face of  the  ground  to  the  horizontal  line  which  passes  through 
the  spirit  level. 

If  strict  accuracy  is  aimed  at,  the  spirit  level  should  be  in 
the  middle  between  the  two  rods.  Considering  D'ED"D  as 
the  spherical  surface  of  the  earth,  and  B'AB"B  as  a  concen- 
tric curve  ;  a  horizontal  line  passing  through  A  is  a  tangent 
to  this  curve.  If  therefore  AT'  and  AT"  are  equal,  the 
points  T'  and  T"  are  equally  distant  from  the  level  of  the 
ocean.  But  if  the  two  rods  are  at  T  and  T',  while  the  spirit 
level  is  at  A,  the  height  TD  is  greater  than  T'D'.  The  dif- 
ference however  will  be  trifling,  if  the  distance  of  the  stations 
T  and  T'  be  small. 

157.  With  these  simple  instruments,  the  spirit  level  and 
the  rods,  the  comparative  heights  of  any  two  places  can  be 
ascertained  by  a  series  of  observations,  without  measuring 
their  distance,  and  however  irregular  may  be  the  ground  be- 
tween them.  But  when  one  of  the  stations  is  visible  from 
the  other,  and  their  distance  is  known  ;  the  difference  of 
their  heights  may  be  found  by  a  single  observation,  provided 
allowance  be  made  for  atmospheric  refraction,  and  for  the 
difference  between  the  true  and  the  apparent  level. 

Problem  I. 

To  find  the  difference  in  the  heights  of  two  places  by    levelling 

rods. 

1 53.  Set  up  the  levelling  rods  perpendicular  to  the  horizon, 
and  at  equal  distances  from  the  spirit  level  ^  observe  the  points 
where  the  line  of  level  strikes  the  rods  before  and  behind,  and 
measure  the  heights  of  these  points  above  the  ground  ;  level  in 
the  same  manner  from  the  second  station  to  the  third,  from  iht 


88  LEVELLING. 

third  to  the  fourth,  <^c.  The  difftrenct  helxveen  the  sum  of  the 
heights  at  the  back  stationSf  and  ai  the  fonoard  stations,  will 
be  the  difference  betiveen  the  height  of  the  first  station  and  the 
last 

If  the  descent  from  H  to  H"(Fig.  49.)  be  required,  let  the 
spirit  level  be  placed  at  A,  equally  distant  from  the  stations 
H  and  H'  ;  observe  where  the  line  of  level  BF  cuts  the  rods 
which  are  at  H  and  li',  and  measure  the  heights  BH  and 
FH'.  The  difference  is  evidently  the  descent  from  the  first 
station  to  the  second.  In  the  same  manner,  by  placing  the 
spirit  level  at  A',  the  descent  from  the  second  station  to  the 
third  may  be  found.  The  back  heights,  as  observed  at  A 
and  A',  are  BH  and  B  H' ;  the  forward  heights  are  FH'  and 
F'H''. 

NowFH'-BH  =the  descent  from  H  to  H', 
And  F'H'  - B'H'=the descent  from  H'  to  H"  ; 
Therefore,  by  addition, 
(FH'-f  F'H'O  -(BH-|-B'H')=the  whole  descent  from  H  to  H". 

159.  It  is  to  be  observed,  that  this  method  gives  the  true 
level,  and  not  the  apparent  level.  The  lines  BF  and  B'F' 
are  not  parallel  to  each  other  ;  but  one  is  parallel  to  a  tan- 
gent to  the  horizon  at  N,  the  other  to  a  tangent  at  N'.  So 
that  the  points  B  and  F  are  equally  distant  from  the  horizon, 
as  are  also  the  points  B'  and  F'.  The  spirit  level  may  be 
placed  at  unequal  distances  from  the  two  station  rods,  if  a 
correction  is  made  for  the  difference  between  the  true  and 
the  apparent  level,  by  problem  IL 

J  60.  If  the  stations  are  numerous,  it  will  be  expedient  to 
place  the  back  and  the  forward  heights  in  separate  columns 
ill  a  table,  as  in  the  following  example. 

Back  heights.     Fore  heights. 


Feet. 

In. 

Feet.  In. 

1st. 

Observation 

3 

7 

2      8 

2. 

2 

5 

3       1 

3. 

6 

3 

5     7 

4. 

4 

2 

3     2 

5. 

5 

9 

4   10 

22 

2 

19     4 

Difference 

19 

4 

2 

10 

LEVELLING.  83 

If  the  sum  of  the  forward  heights  is  less  than  the  sum  of 
ihe  back  heights,  it  is  evident  that  the  last  station  must  bp 
higher  than  the  first. 

Problem  II. 

161.  To  find  the  difference  between  the  true  and  the  ap- 
parent level,  for  any  given  distance. 

If  C  (Fig.  12.)  be  the  centre   of  the  earth  considered  as  a 
sphere,  AB  a  portion  of  its  surface,  and  T  a  point  on  an  ap- 
parent level   with  A  ;  then  BT  is  the  difference  between  the 
true  and  the  apparent  level,  for  the  distance  AT. 
Let  2BC  =  D,  the  diameter  of  the  earth, 

AT=c/,  the  distance  of  T,  in  a  right  line,  from  A, 
BT=A,  the  height  of  T,  or  the  difference  between 
the  true  and  the  apparent  level. 

Then  by  Euc.  36.  3, 
(2BC  +  BT)XBT=AT';  that  is,  {D+h)xh==d^ 
and  reducing  the  equation, 

Therefore,  to  find  h  the  difference  between  the  true  and 
the  apparent  level,  add  together  one  fourth  of  the  square  of 
the  earth's  diameter,  and  the  square  of  the  distance,  extract 
the  square  root  of  the  sum,  and  subtract  the  semi-diameter 
of  the  earth. 

162.  This  rule  is  exact.  But  there  is  a  more  simple  one, 
which  is  sutficiently  near  the  truth  for  the  common  purposes 
of  levelling.  The  height  BT  is  so  small,  compared  with  the 
diameter  of  the  earth,  that  D  may  be  substituted  for  D-f^, 
without  any  considerable  errour.  The  original  equation 
above  will  then  become 

Dxh=dK     Therefore  h=^— 

D 

That  is,  the  difference  between  the  true  and  the  apparent 
level,  is  nearly  equal  to  the  square  of  the  distance  divided  by 
*he  diameter  of  the  earth. 

Ex.  I.  What  is  the  difference  between  the  true  and  the 
app.  nt  level,  for  a  distance  of  one  English  mile,  supposing 
the  earth  to  be  7940  miles  in  diameter  ? 

Ans.  7.98  inches,  or  8  inches  nearly* 
13 


90  SURVEYING.  , 

In  the  equation  ^=7^-*  as  D  is  a   constant  quantity,  it   is 

evident  that  Aw  varies  as  c?*.  According  to  the  last  rule  then, 
the  differenee  hetween  the  true  and  the  apparent  level  varies 
as  the  square  of  the  distance.  The  difference  for  1  mile  being 
nearly  8  inches, 

In.      Feet.  In 

For  2  miles,  it  is  8X22  =  32=2     8  nearly. 
For  3  miles,  8X3^=  G 

For  4  miles,         8X43=  jq  g 

&c.  &c.     See  Table  IV. 

Ex.  2.  An  observation  is  made  to  determine  whether  wa- 
ter can  be  brought  into  a  town  from  a  spring  on  a  neighbour- 
ing hill.  At  a  particular  spot  in  the  town,  the  spring,  which 
is  2i  miles  distant,  is  observed  to  be  apparently  on  a  level. 
What  is  the  descent  from  the  spring  to  this  spot  ? 

The  descent  is  nearly  4  feet  2  inches  for  the  whole  dis- 
tance, or  20  inches  in  a  mile;  which  is  more  than  sufficient 
for  the  water  to  run  freely. 

Ex.  3.  A  tangent  to  a  certain  point  on  the  ocean,  strikes 
the  top  of  a  mountain  23  miles  distant.  What  is  the  height 
of  the  mountain  ?  Ans.  352  (eet, 

1G3.  One  place  may  be  below  the  apparent  level  of  an- 
other, and  yet  above  the  true  level.  The  difference  between 
the  true  and  the  apparent  level  for  3  miles  is  6  feet.  If  one 
spot,  then,  be  only  two  feet  below  the  apparent  level  of  an- 
other 3  miles  distant,  it  will  really  be  4  feet  higher. 

If  two  places  are  on  the  same  true  level,  it  is  evident  that 
each  is  below  the  apparent  level  of  the  other. 

Problem  III. 

To  Jind  the  difference  in  the  heights  of  two  places  whose  dis- 
tance is  known, 

1G4.  From  the  angle  of  elevation  or  depression,  calculate 
how  far  one  of  tie  places  is  above  or  below  the  apparent  level 
of  the  other  ;  and  then  make  allowance  for  the  difference  be- 
tween  the  apparent  and  the  true  level. 

By  taking,  with  a  quadrant,  the  elevation  of  the  object 
whose  distance  is  given,  we  have  one  side  and  the  angles  of 


LEVELLING.  91 

a  fight  angled  triangle,  to  find  the  perpendicular  height  above 
a  horizontal  plane.  (Art.  6.)  Adding  this  to  the  difference 
between  the  true  and  the  apparent  level,  we  have  the  height 
of  the  object  above  the  true  level  of  the  place  of  observation. 
Whi'n  an  angle  o(  depression  is  taken,  it  will  be  necessary  to 
subtract  instead  of  a<lding. 

Ex.  L  The  angle  of  elevation  of  a  hill,  as  observed  from 
the  top  of  another  44^  miles  distant,  is  found  to  be  7  degrees. 
AVhat  is  the  difference  in  the  heights  of  the  two  hills  ? 

Heij^ht  of  one  above  the  level  of  the  other  2917.3  feet. 

Difference  of  the  levels  13.5 


Difference  in  the  heights  of  the  hills  2930.8 

Ex.  2.  From  the  top  of  a  tower,  the  angle  of  depression 
of  a  fort  4  miles  distant,  is  found  to  be  3  J  degrees.  What 
is  the  height  of  the  tower  above  the  fort  ? 

Ans.  1189  feet. 

if  the  operation  of  levelling  is  meant  to  be  very  exact,  es- 
pecially when  extended  to  considerable  distances,  allowance 
should  be  made  for  atmospheric  refraction,* 


»  Sec  Nol«  A, 


SECTION  T. 

THE  MAGNETIC  NEEDLE.* 


A          ifir    T^H^  direction  in  which  a  ship  is  steered,  and 
^*  ■    -*-    the  bearings  of  the  sides  of  a  field,  are  com- 

monly determined  by  observing  the  angles  which  ihey  make 
with  the  magnetic  needle.  This  is  a  bar  of  steel  to  which 
the  magnetic  power  has  b(^en  communicated  from  some  other 
artificial  or  natural  magnet.  When  it  is  balanced  on  a  pin, 
so  as  to  turn  freely  in  any  direction,  it  points  towards  the 
north  and  south. 

The  poles  of  the  needle  are  its  two  extremities  ;  and  the 
vertical  plane  which  passes  through  these,  is  called  the  ma^- 
netic  meridian.  The  astronomical  meridian  passes  through 
the  poles  of  the  earth.  These  two  meridians  rarely  coin- 
cide. The  needle  does  not  often  point  directly  north  and 
south. 

167.  The  DECLINATION  of  the  needle  is  the  angle  which  it 
77iakes  with  a  north  and  south  line  ;  or  the  angle  between  the 
magnetic  and  the  astronomical  meridians.  It  is  said  to  be 
castor  west,  according  as  the  north  pole  of  the  needle  points 
east  or  west  of  the  north  pole  of  the  earth. 

The  variation  of  the  needle  is  properly  the  change  of  its 
declination.  The  term,  however,  is  frequently  used  to  signi- 
fy the  declination  itself. 

The  declination  of  the  needle  is  very  different  in  different 
parts  of  the  earth.  In  some  places,  it  is  20  or  30  degrees  ;^ 
in  others,  little  or  nothing.  In  the  variation  charts  given  by 
writers  on  magnetism,  the  declination  is  marked,  as  it  is 
found  by  observation  on  different  parts  of  the  globe.  Lines 
are  drawn  connecting  all  the  points  which  have  the  same  de- 
clination. Thus  a  line  is  drawn  through  the  several  places 
in  which  the  declination  is  10  degrees  ;  another  through  those 


*  Cavallo  on  Magnetism,  Rees*  Cyclopedia,  Transactions  of  the  Royal 
Society  of  London,  the  Royal  Irish  Academy,  the  American  Philosophical 
•Society,  end  the  American  Academy  of  Arts  and  Sciences. 


THE  MAGNETIC  NEEDLE.  m 

in  which  it  is  5  degrees,  &c.  These  lines  are  very  winding  ; 
yet  they  never  cross  each  other,  though  they  extend  all  over 
the  globe.  One  of  the  lines  of  wo  declination  passes  through 
the  middle  parts  of  the  United  States.  The  declination  is 
towards  this  line,  in  places  which  are  on  either  side  of  it. 
Thus  in  New-England  the  declination  is  west,  while  on  the 
Ohio  it  is  east.  It  increases  with  the  distance  from  the  line 
of  no  declination. 

168.  The  declination  is  not  only  different  in  different  pla- 
ces, but  different  in  the  same  place  at  different  times.  At 
London,  about  230  years  since,  it  was  11^-  degrees  east.  It 
gradually  decreased  till  1657,  when  the  needle  pointed  di- 
rectly north  From  that  time  it  deviated  more  and  more  to 
the  west,  till  in  1800  the  declination  became  about  24  de- 
grees. At  present,  it  appears  to  be  nearly  stationary,  both 
at  London  and  Paris. 

In  New-England,  the  declination  h^s  been  generally  de- 
creasing, for  many  years.  At  Boston,  it  was  about  9  degrees 
in  1 708,  8  degrees  in  1 742,  7  degrees  in  1 782,  and  6^  degrees 
in  1810;  the  rate  of  variation  being  about  H'  in  a  year,  or  a 
deforce  in  40  years.* 

The  variation  in  the  declination  is  by  no  means  uniform. 
If  the  needle  moves  two  minutes  from  the  meridian  in  one 
year,  it  may  move  a  greater  or  less  distance  the  next  year 
its  progress  is  different  in  different  places.  In  some  it  ig 
moving  east,  and  in  others  west  ;  in  some  it  is  coming  near- 
er to  the  meridian,  in  others  going  farther  from  it. 

169.  There  is  also  a  c^mrna/ variation,  which  appears  to 
be  owing  to  a  change  of  temperature.  During  the  fore  part 
of  the  day,  the  north  end  of  the  needle  frequently  moves  a 
few  minutes  of  a  degree  to  the  west.  In  the  evening,  it  re- 
turns nearly  to  the  same  point  from  which  it  started.  This 
diurnal  variation  is  found  to  be  the  greatest  in  the  summer 
months,  when  the  action  of  the  sun  is  the  most  powerful. 

In  addition  to  these  various  changes,  there  are  local  per- 
turbations of  the  needle,  occasioned  probably  by  the  attrac- 
tion of  ferruginous  substances  beneath  the  surface  of  the 
ground. 


*  See  the  obserrations  of  Mr.  Bowditch,  in  the  Memoirs  of  the  American 
Academy  of  Arts  and  Sciences,  Vol.  III.  twi  II.  p.  337. 


94  THE  MAGNETIC  NEEDLE. 

170.  So  many  irregularities  must  render  the  magnetic 
compass  an  inaccurate  instrument,  unless  the  state  of  the  de- 
clination is  ascertained  by  frequent  observations.  This  is 
particularly  necessary  at  sea,  where  the  declination  may  be 
changed  by  a  few  hours  sail. 

The  astronomical  meridian  is  determined  by  the  positions 
of  the  heavenly  bodies.  The  situation  of  the  sun  at  rising 
or  setting  being  known,  its  distance  from  the  magnetic  me- 
ridian may  be  observed  with  an  azimuth  compass,  which  is  a 
mariner's  compass  with  the  addition  of  sight  vanes  for  taking 
the  direction  of  any  object. 

17i.  On  land,  a  true  meridian  line  may  be  drawn  by  ob- 
servations on  ihepole  star.  If  this  were  exactly  in  the  pole, 
it  would  be  always  on  the  meridian.  But  the  star  revolves 
round  the  pole,  at  a  short  distance,  in  a  little  less  than  24 
hours.  In  about  6  hours  from  its  passing  the  meridian  above 
the  pole,  it  is  at  its  grea^test  distance  west ;  in  about  12  hours, 
it  is  on  the  meridian  beneath  the  pole,  and  in  about  1 8  hours, 
at  its  greatest  distance  east.  If  the  direction  of  the  star  can 
be  taken,  at  the  instant  when  it  is  on  the  meridian,  either 
above  or  beneath  the  pole  ;  a  true  north  and  south  line  may 
be  found.  This  method,  however,  requires  that  the  exact 
time  of  passing  the  meridian  be  known,  and  that  the  obser- 
vations be  made  expeditiously. 

172.  But  as  the  star  comes  very  gradually  to  its  greatest 
distance  east  or  west,  it  is  easy  to  observe  these  limits  ;  and 
as  the  revolution  is  made  in  a  circle  round  the  axis  of  the 
earth,  it  is  evident  that  the  pole  must  be  in  the  middle  be- 
tween the  two  extreme  distances.  To  draw  a  true  meridian 
line,  then,  take  the  direction  of  the  pole  star  when  it  is  farthest 
west,  and  also  when  it  is  farthest  east  ;  and  bisect  the  angle 
made  by  these  two  directions. 

When  a  meridian  is  once  drawn,  it  may  be  rendered  per- 
manent, by  fixing  proper  marks  and  the  declination  of  the 
needle  may  then  be  ascertained  at  any  time,  by  the  survey- 
or's compass,  or  more  accurately  by  the  variation  compass ^ 
which  has  a  long  needle,  and  a  graduated  arc  of  so  large  a 
radius  as  to  admit  of  very  accurate  divisions.* 

*  See  Note  L. 


NOTES. 


Note  A.     Page  10  and  91. 


A  Ray  of  light,  in  coming  from  a  distant  object  to  tlu 
eye,  through  the  air,  is  turned  from  a  straight  line  into  a 
curve  which  is  concave  towards  the  earth.  The  effect  is  to 
e/era^e  the  apparent  place  of  the  object,  as  each  point  ap- 
pears in  the  direction  in  which  the  light  from  that  point  en- 
ters the  eye.  This  change  in  the  apparent  situation  is  call- 
ed astronomical  refraction,  when  the  heavenly  bodies  are 
concerned  ;  and  terrestrial  refraction,  when  the  objects  are 
on  the  earth.  The  measure  of  the  latter  is  the  angle  at  the 
eye,  between  a  straight  line  drawn  to  the  object,  and  a  tan* 
gent  to  the  curvilinear  ray,  as  TAT',  (Fig.  50.)  T  being  the 
place  of  the  object,  and  T'  its  apparent  place  as  seen  from 
A. 

The  refraction  is  very  much  affected  by  the  state  of  the 
atmosphere  ;  changing  with  the  temperature,  as  well  as  with 
the  density  indicated  ny  the  barometer.  Jn  the  delicate  ob- 
servations made  in  the  trigonometrical  surveys  in  England 
and  France,  the  terrestrial  refracti  n  was  found  to  vary  from 
I  to  ^\  of  the  angle  at  the  centre  of  the  earth  subtended  by 
the  distance  of  the  object.  The  mean  is  y'^  :  thus  if  an  ob- 
ject at  T  (Fig.  50.)  as  seen  from  A  in  the  mean  state  of  the 
atmosphere,  appears  to  be  raised  to  T' ;  the  angle  TAT'  is 
about  j\  of  the  angle  ACT  subtended  by  the  distance  AT. 
This  angle  is  easily  found  from  the  arc  AB,  which  is  nearly 
equal  to  AT  ;  the  whole  circumference  of  the  earth  being  to 
the  arc,  as  360  degrees  to  the  angle  required.  The  mean 
terrestrial  refraction,  as  thus  calculated,  is  3'.  7  for  a  mile, 
;4nd  increases  as  the  distance  nearly  ;  the  elevation  of  the 
object  being  supposed  to  be  small  in  comparison  with  its  dis- 
tance. In  measuring  altitudes,  the  terrestrial  refraction  is 
to  be  subtracted  from  the  observed  angle  of  elevation. 


9tJ  NAVIGATION. 

The  alteration  in  the  height  of  the  object,  by  the  mean  re- 
fraction, is  equal  to  j  of  the  curvature  of  the  earth  for  the 
given  distance,  or  of  the  difference  between  the  true  and  ap- 
parent levels,  as  calculated  by  the  rule  in  Art.  162.  If  the 
angle  of  refraction  were  equal  to  ha/f  the  angle  at  the  centre 
of  the  earth  subtended  by  the  distance,  the  change  in  the 
height  of  the  object  would  be  just  equal  to  the  correction  for 
the  curvature.  If  an  object  at  B  (Fig.  12.)  were  raised  by 
refraction,  so  as  to  be  seen  from  A  in  the  direction  of  the  tan- 
gent AT  ;  the  change  in  the  altitude  would  be  equal  to  BT, 
which  is  the  difference  between  the  true  and  the  apparent 
level  of  A.  In  this  case,  the  angle  BAT  would  be  half  ACB. 
(Euc.  32.  3  and  20.  3.)  But  as  the  angle  of  refraction  is  in 
fact  only  y'^  of  the  angle  at  the  centre,  the  change  in  the  al- 
titude is  only  4  of  the  correction  for  curvature.  The  latter 
is  about  f  of  a  foot  for  a  mile,  and  varies  as  the  square  of  the 
distance.  If  then  d  be  the  distance  in  miles  ;  the  correction 
for  the  curvature  will  be  fc?^,  and  the  correction  for  refrac- 
tion ^^^^^     See  Table  IV. 

The  greatest  distance  at  which  an  object  can  be  seen  on 
the  surface  of  the  earth,  as  calculated  by  the  rule  in  Art.  23, 
depends  on  the  apparent  altitude.  This  being  to  the  real  al- 
titude as  7  to  6,  and  the  distance  being  nearly  as  the  square 
root  of  the  altitude  ;  the  distance  at  which  an  object  can  be 
seen  by  the  mean  refraction,  is  to  the  distance  at  which  it 
could  be  seen  without  refraction  as  \/7  :  \/6,  or  as  14  :  13 
nearly.  See  Playfair's  Astronomy,  Sec.  II.  Vince's  Astron- 
omy, Chap.  VII.  and  the  accounts  of  the  Trigonometrical 
Surveys  in  England  and  France. 


Note  B.  p.  20. 


The  method  of  calculation  in  plane  sailing  is  sometimes 
spoken  of  as  inaccurate,  as  only  approximating  to  the  truth, 
in  proportion  to  the  smallness  of  the  difference  between  a 
plane  and  that  part  of  the  ocean  to  which  the  calculation  is 
applied.  This  view  of  the  subject  appears  not  to  be  strictly 
correct.  It  is  true,  that  plane  sailing  is  incomplete,  as  it  does 
not  ascertain  the  longitude.  This  belongs  to  middle  latitude 
or  Mercator's  sailing.  It  is  also  true,  that  if  a  ship  sails  on 
several  courses^  the  sum  of  the  departures  is  not  equal  to  the 


NOTES.  d7 

departure  for  the  same  distance  on  a  single  course,  as  would 
be  th('  case  on  a  plane.  (Art.  78.)  It  is  farther  to  be  ob- 
served, that  the  departure,  as  calculated  by  plane  sailing,  is 
neither  the  meridian  distance  measured  on  the  parallel  of 
latitude  from  which  the  ship  sails,  nor  that  measured  upon 
the  parallel  upon  which  she  arrives.  But  the  departure  for 
a  single  course,  as  defined  in  Art.  40,  and  the  ditFerence  of 
latitude,  are  as  accurately  calculated  by  plane  sailing,  as  if 
the  surface  of  the  ocean  were  a  plane.  Let  the  whole  dis- 
tance be  divided  into  portions  so  small,  that  one  of  the  arcs 
shall  differ  less  from  its  tangent  than  by  any  given  quantity. 
Each  of  these  portions  is  to  the  corresponding:  departure,  as 
radius  to  the  sine  of  the  course  ;  and  to  the  difference  of  lati- 
tude, as  radius  to  the  cosine  of  the  course.  Therefore  the 
whole  distance  is  to  the  whole  departure,  as  radius  to  the 
sine  of  the  course  ;  and  to  the  whole  difference  of  latitude, 
as  radius  to  the  cosine  of  the  course.  These  proportions 
are  exact,  even  for  a  spheroid,  a  cylinder,  or  any  solid  of 
revolution. 

If  there  were  any  incorrectness  in  plane  sailing,  it  would 
extend  to  Mcrcator's  sailing  also  ;  for  one  is  founded  on  the 
other.  In  Mercator's  sailing,  the  proper  difference  of  lati- 
tude is  to  the  meridional  difference  of  latitude,  as  the  depar- 
ture to  the  difference  of  longitude.  Now  the  departure  is 
calculated  by  plane  sailing  ;  and  any  errour  in  this  must  pro- 
duce an  errour  in  the  longitude.  Or  if  the  longitude  be 
found  by  the  theorem  in  Art.  72,  without  previously  calcu- 
lating the  departure  •,  yet  the  table  of  meridional  parts  which 
must  be  used,  is  founded  on  the  ratio  between  the  departure 
and  the  difference  of  longitude.     Art.  G5. 


Note  C.  p.  27. 


It  is  here  supposed  that  the  direction  of  the  ship  is  at 
right  angles  with  every  meridian  which  she  crosses.  A  num- 
ber of  curious  questions  have  been  started  respecting  sailing 
on  a  sphere  :  such  as  whether  a  due  east  and  west  line  coin- 
cides with  a  parallel  of  latitude,  &z;c.  Most  of  these  points 
are  easily  settled  by  proper  definitions.  But  this  is  not  the 
place  to  consider  them,  as  they  belong  to  spherical  ge- 
ometry. 

14 


9H  NAVIGATION. 

Note  D.  p.  34. 

As  the  length  of  a  minute  of  Mercator's  meridian,  is  equal 
to  the  secant  of  the  latitude,  it  will  be  a  little  more  exact  to 
take  the  latitude  of  the  middle  of  the  arc,  rather  than  that  of 
one  extremity.  On  the  extended  meridian,  the  first  minute 
will  then  be  made  equal  to  the  secant  of  ^',  the  second  to 
the  secant  of  H',  the  third  to  the  secant  of  2^',  (fee. 

The  method  of  calculating  by  natural  secants,  though  use- 
ful in  forming  a  table  of  meridional  parts,  is  subject  to  this 
inconvenience,  that  to  obtain  the  meridional  parts  for  any 
number  of  degrees  of  latitude,  it  is  necessary  to  find  sepa- 
rately the  parts  for  each  of  the  minutes  contained  in  the 
given  arc,  and  then  to  add  them  together.  There  is  a  dif- 
ferent method,  by  which  the  meridional  parts  for  an  arc  of 
any  extent,  may  be  calculated  independently  of  any  other 
arc.  A  portion  of  Mercator's  meridian,  extending  from  the 
equator  to  a  given  latitude,  the  semi-diameter  of  the  earth 
being  1,  is  equal  to  the  hyperbolic  logarithm  of  the  cotangent 
of  half  the  complement  of  the  latitude.  See  the  London  Phi- 
losophical Transactions,  vol.  xix.  No.  219,  Vince's  Fluxions, 
Art.  190,  and  the  Introduction  to  Hutton's  Mathematical 
Tables. 

Note  E.  p.  39. 

The  distance  which  a  ship  sails,  in  going  from  one  place 
to  another  on  a  rhumb  line,  is  not  the  nearest  distance  ;  for 
this  would  be  an  arc  of  a  great  circle.  To  sail  on  a  great 
circle,  except  on  a  meridian  or  the  equator,  she  must  be  con- 
tinually altering  her  course.  If  it  were  practicable  to  steer 
a  vessel  in  this  manner,  the  departure,  difference  of  latitude, 
&c.  might  be  calculated  by  spherical  trigonometry. 

Note  F.  p.  41. 

A  traverse  may  also  be  constructed  like  the  plot  of  a  field 
in  surveying,  either  by  drawing  parallel  lines,  as  in  Art.  Ill, 
or  from  the  angles  given  by  the  rules  in  Art.  112,  or  more 
simply,  as  in  Art.  1 23,  by  departure  and  difference  of  latitude, 
when  these  have  been  found  by  calculation  or  inspection. 


NOTES.  99 


Note  G.  p.  46. 


Plane  sailing  is  sometimes  represented  as  a  method  of 
calculation  founded  on  the  principles  of  the  plane  chart.  But 
in  the  construction  of  this  chart,  a  principle  is  assumed  which 
is  known  to  be  erroneous.  That  part  of  the  surface  of  the 
earth  which  is  represented  on  it,  is  supposed  to  be  a  plane. 
This  renders  the  construction  more  or  less  inaccurate.  But 
in  plane  sailing,  the  calculations  are  strictly  correct.  The 
principle  assumed  here  is  not  that  the  surface  of  the  earth 
is  a  plane;  but  that,  from  the  peculiar  nature  of  the  rhumb 
line,  the  distance,  departure,  and  difference  of  latitude,  where 
the  course  is  given,  have  the  same  ratio  to  each  other  which 
they  would  have  upon  a  plane. 


Note  H.  p.  51. 


The  Quadrant  of  reflexion  has  received  the  name  of  Had' 
tey^s  Quadrant^  as  the  description  of  it  which  was  6rst  made 
public,  was  given  by  John  Hadley,  Esq.  But  he  has  not  an 
undisputed  claim  to  the  first  invention.  His  description  of 
the  instrument  was  communicated  to  the  Royal  Society  of 
London,  in  May,  1731.  It  appears  that  the  principle  on 
which  it  is  constructed  had  been  suggested  by  Dr.  Hooke, 
several  years  before.  But  the  form  which  he  proposed  was 
not  calculated  to  answer  the  purpose,  as  it  admitted  of  only 
one  reflexion.  Sir  Isaac  Newton,  however,  who  died  in 
1727,  left  among  his  papers  a  description  of  a  quadrant  with 
two  reflexions,  which  is  substantially  the  same  as  Hadley's. 
This  was  published  in  the  Philosophical  Transactions  for 
1742. 

It  is  also  stated  that  a  quadrant  similar  to  Hadley's  had 
been  contrived  by  Mr.  Thomas  Godfrey,  of  Philadelphia, 
before  Hadley's  description  was  communicated  to  the  Royal 
Society. 

Hooke's  Posthumous  Works,  Hutton's  Dictionary,  Trans- 
actions of  the  Royal  Society  of  London  for  1731,  1734,  and 
1742,  American  Magazine  for  Aug.  and  Sept,  1758,  Millar's 
Retrospect,  i.  p.  468,  and  Analectic  Magazine,  ix.  281. 


!0«  NAVIGATION. 


Note  I.  p.  56. 

The  proportion  in  Art.  99,  on  account  of  the  smallness  of 
the  height  BT  compared  with  the  semi-diameter  of  the 
earth,  is  not  very  suitable  for  calculating  the  depression  with 
exactness.  The  following  rule,  which  includes  the  effect  of 
refraction,  is  better  adapted  to  the  purpose.  The  depres- 
sion is  found  by  multiplying  59"  into  the  square  root  of  the 
height  in  feet.  See  Vince's  Astronomy,  Art.  197,  Rees' Cy- 
clopedia, and  Table  II. 

In  taking  the  altitude  of  a  heavenly  body  with  Hadley's 
Quadrant,  when  the  view  of  the  ocean  is  unobstructed,  the 
reflected  image  is  made  to  coincide  with  the  most  remote 
visible  point  of  the  water.  But  when  there  is  land  in  the 
direction  in  which  the  observation  is  to  be  made,  the  image 
is  brou;;ht  to  the  water's  edge  ;  and  the  dip  is  increased,  in 
proportion  as  the  distance  of  the  land  is  diminished.  See 
table  III. 


Note  K.  p.  81. 

This  is  not  the  place  for  a  detailed  account  of  the  various 
trigonometrical  operations  which  have  been  undertaken,  for 
the  purpose  of  determining  the  length  of  a  degree  of  lati- 
tude, in  different  parts  of  the  earth.  The  subject  belongs 
rather  to  astronomy,  than  to  common  surveying.  It  may  not 
be  amiss,  however,  to  give  a  concise  statement  of  the  meas- 
urements which  have  been  made  in  the  present  and  the  pre- 
ceding century. 

About  the  year  1700,  Picard  measured  a  degree  between 
Paris  and  Amiens  ;  and  the  arc  was  extended  by  Cassini  to 
Perpignan,  about  6  degrees  south  of  Paris,  and  afterwards  to 
the  northward  as  far  as  Dunkirk.  These  measurements  were 
made  in  the  middle  latitudes.  To  compare  a  degree  here, 
with  the  length  of  one  near  the  pole,  and  another  on  the 
equator,  two  expeditions  were  fitted  out  from  France  about 
tVie  same  time,  one  for  Lapland,  and  the  other  for  South- 
America.  The  latter  sailed  in  May,  1735,  for  Peru,  and  af- 
ter a  series  of  the  most  formidable  embarrassments,  they 
succeeded,  at  the  end  of  0  years,  in  accomplishing  their  ob- 
ject. They  measured  an  arc  of  the  meridian,  crossing  the 
equator  from   3®  7'  north  latitude  to  about  3^**  south.     The 


NOTES.  161 

other  party,  in  1736,  under  the  direction  ot^Maupertius,  pro- 
ceeded to  the  head  of  the  gulf  of  Bothnia,  and  measured 
an  arc  of  the  meridian  extending  along  the  river  Tornei,  and 
crossing  the  polar  circle.  The  difficulties  which  they  expe- 
rienced, in  this  frozen  and  desolate  region,  were  scarcely  in- 
feriour  to  those  with  which  the  other  adventurers  were  at  the 
same  time  contending  in  South-America. 

The  determination  of  these  three  arcs,  one  in  France,  one 
in  Peru,  and  one  in  Lapland,  were  sufficient  to  satisfy  astro- 
nomers, that  a  degree  of  latitude  near  the  poles  is  greater 
than  one  on  the  equator,  and  consequently  that  the  equatorial 
diameter  of  the  earth  is  longer  than  the  polar.  But  it  does 
Bot  necessarily  follow  from  this,  that  there  is  a  regular  in- 
crease in  the  length  of  a  degree,  from  the  lower  to  the  high- 
er latitudes.  On  the  contrary,  according  to  the  survey  which 
had  been  made  by  Picard  and  others,  a  degree  was  found  to 
he  greater  in  the  south  of  France,  than  in  the  north.  The 
zeal  of  astronomers  was  therefore  excited  to  take  farther 
measures,  to  determine  what  is  the  exact  length  of  a  degree, 
in  various  parts  of  the  earth,  and  to  ascertain  whether  in  the 
influence  of  gravitation,  there  are  local  inequalities,  which 
affect  the  astronomical  observations. 

La  Caille,  about  this  time,  measured  an  arc  of  the  meri- 
dian at  the  Cape  of  Good  Hope.  He  was  not,  however, 
provided  with  such  instruments  as  would  insure  a  great  de- 
gree of  precision.  Boscovich,  a  distinguished  philosopher, 
measured  an  arc  of  two  degrees  in  Italy,  from  Rimini  to 
Rome.  Between  the  years  1764  and  1768,  Messrs.  Mason 
and  Dixon,  under  the  direction  of  the  Royal  Society  of  Lon- 
don, measured  an  arc  of  the  meridian  of  about  one  degree 
and  an  half,  crossing  the  line  betw^een  Pennsylvania  and  Ma- 
ryland. As  the  country  here  is  very  level,  the  whole  dis- 
tance was  measured,  not  by  a  combination  of  triangles,  but 
in  the  first  place  with  a  chain,  and  afterwards  with  rods  of 
fir.  A  degree  was  also  measured  in  Piedmont,  another  in 
Austria,  and  a  third  in  Hungary  ;  the  first  by  Beccaria,  and 
the  two  latter  by  Liesganig. 

But  the  most  perfect  of  all  the  trigonometrical  surveys  up- 
on a  great  scale,  are  those  which  have  been  made  within  a 
few  years  in  England  and  France.  The  instruments  used 
for  taking  the  angles,  particularly  the  theodolite  of  Ramsden, 
and  the  repeating  circle  of  Borda,  have  been  brought  to  a 
surprising  degree  of  exactness.     The  re-measurement  of  the 


102  SURVEYING. 

line  from  Dunkirk  to  Barcelona,  a  part  of  which  had  been 
several  times  surveyed  before,  was  commenced  in  1792,  un- 
der the  direction  of  the  Academy  of  Sciences  in  Paris;  for 
the  purpose  of  obtaining  a  standard  of  measure  of  lengths, 
weights,  capacity,  <Sz;c.  derived  from  a  portion  of  the  meridi- 
an. The  northern  part  of  the  arc  was  measured  by  Delam- 
bre,  and  the  southern  part  by  Mechain,  who  lost  his  life  in 
1805,  in  attempting  to  extend  the  line  beyond  Barcelona,  to 
the  Balearic  Islands  in  the  Mediterranean.  This  line  was  af- 
terwards continued  by  13iot  to  Formentera,  the  southernmost 
of  these  islands,  which  is  in  Lat.  38°  38'  56".  The  latitude 
of  Dunkirk  is  51°  2'  9".  The  whole  arc  is,  therefore,  more 
than  12  degrees,  and  is  nearly  bisected  by  the  45th  parallel 
of  latitude.  This  is  connected  with  the  line  carried  through 
England  to  Clifton  in  Lat.  53°  27'  31'':  making  the  whole 
extent  nearly  15  degrees. 

The  arc  which  had  been  surveyed  by  Maupertius,  on  the 
polar  circle,  was  re-measured  by  Swanberg  and  others,  in 
1802.  There  is  a  difference  of  230toise3  in  the  length  of  a 
degree,  as  calculated  from  these  two  measurements.  In  In- 
dia, an  arc  of  the  meridian  was  measured,  on  the  coast  of 
Coromandel,  in  1803,  by  Major  Lambton. 

According  to  these  various  measurements,  we  have  the 
following  lengths  of  a  degree  of  latitude  in  different  parts  of 
the  earth. 

Latitude.     Toiees.    Fathoms. 

1 .  In  Peru,  by  Bouguer,  Q.  56,750    60,480 

2.  In  India,  by  Lambton,  11°  30*    56,756    60,487 

3.  AttheCap^jfGoodHope,|       33^3     57^037    60,780 

4.  In  Pennsylvania, 'by  Mason)       33  ^^     56,890    60,630 

and  Dixon,                        )                         '  ' 

5.  In  Italy,  by  Boscovich,  43  56,980  60,725 
a.  In  Piedmont,  by  Beccaria,  44  44  57,070  60,820 
7.  InFrance,byDelambreand>  ^^  ^^^  ^^^^^ 

Mecham,  5 

3.   In  Austria,  by  Liesganig,                48  43  57,086  60,835 

6.  In  England,  by  Roy  and  Mudge,  52  2  57,074  60,827 
10.  In  Lapland,  by  Maupertius,        66  20  57,422  61,184 

Do.  by  Swanberg,  57,192    60,952 

On  a  comparison  of  all   the    measurements   which  have 

been  made,  it  is  found  that  a  degree  of  latitude  is  greater 

near  the  poles,  than  in  the  middle  latitudes  ;  and  greater  in 

ttie  middle  latitudes,  than  near  the   equator.     The  earth  is 


NOTES.  103 

therefore  coBipressed  at  the  poles,  and  extended  at  the  equa- 
tor. But  it  does  not  appear  that  it  is  an  exact  spheroid,  or 
a  solid  of  revolution  of  any  kind.  If  arcs  of  the  meridian 
which  are  near  to  each  other  and  of  moderate  length  be 
compared,  they  will  not  be  found  to  increase  regularly  from 
a  lower  to  a  higher  latitude.  On  the  southern  part  of  the 
line  which  was  measured  in  France,  the  degrees  increase  ve- 
ry slowly  ;  towards  the  middle,  very  rapidly ;  and  near  the 
northern  extremity,  very  slowly  again.  Similar  irregulari- 
ties are  found  in  that  part  of  the  meridian  which  passes 
through  England.  These  irregularities  are  too  great  to  be 
ascribed  to  errours  in  the  surveys.  It  is  concluded,  there- 
fore, that  the  direction  of  the  plumb  line,  which  is  used  in 
determining  the  latitude,  is  affected  by  local  inequalities  in 
the  action  of  gravitation,  owing  probably  to  the  different  den- 
sities of  the  substances  of  which  the  earth  is  composed. 
These  inequalities  must  also  have  an  influence  upon  the  fig- 
ure of  the  fluid  parts  of  the  globe,  so  that  the  surface  ought 
not  to  be  considered  as  exactly  spheroidical. 

See  Col.  Mudge's  account  of  the  Trigonometrical  Survey 
in  England.  Gregory's  Dissertations,  &c.  on  the  Trigono- 
metrical Survey.  Rees' Cyclopedia,  Art.  Degree.  Playfair's 
Astronomy.  Philosophical  Transactions  of  London  for  17G8, 
1785,1787,1790,1791,1795,1797,  1800.  Asiatic  Research- 
es, vol.  viii.  Puissant.  "  Traite  de  Geodesic."  Maupertius. 
"  Degre  du  Meridian  entre  Paris  et  Amiens."  Do.  "  La- 
Figure  de  la  Terre.''  Cassini.  "  Expose  des  Operations, 
&ic."  Delambre.  *' Bases  du  systeme  m^trique.''  Swanberg. 
"  Exposition  des  Operations  faites  en  Lapponie."  Laplace. 
"  Trait*  de  Mecanique  Celeste." 


Note  L.  p.  94. 

One  of  the  most  simple  methods  of  determining  when  the 
pole  star  is  on  the  meridian,  is  from  the  situations  of  two 
other  stars,  Alioth  and  y  Cassiopeiae,  both  which  come  on  to 
the  meridian  a  few  minutes  before  the  pole  star,  the  one 
above  and  the  other  below  the  pole.  Alioth,  which  is  the 
star  marked  sin  the  Great  Bear,  is  on  the  same  side  of  the 
pole  with  the  pole  star,  and  about  30  degrees  distant.  The 
star  7  in  the  constellation  Cassiopeiae,  is  nearly  as  far  on  the 


104  SURVEYING. 

opposite  side  of  the  pole.  The  right  ascension  ot*  the  latter  in 
1810  was  Oh.  45m.  24s.,  increasing  about  3|  seconds  an- 
nually. The  right  ascension  of  Alioth  was  12h.  45m.  36s., 
increasing  about  2i  seconds  annually.  These  two  stars, 
therefore,  come  on  to  the  meridian  nearly  at  the  same  time^ 
This  time  may  be  known  by  observing  when  the  same  verti- 
cal line  passes  through  them  both.  The  right  ascension  of 
the  pole  star  in  January  1810,  was  Oh.  54m.  36s.,  and  in- 
creases 13  or  14  seconds  in  a  year.  So  that  this  star  comes 
to  the  meridian  about  9  or  10  minutes  after  /  Cassiopeiae. 
In  very  nice  observations,  it  will  be  necessary  to  make  allow- 
ance for  nutation,  aberration,  and  the  annual  variation  in 
right  ascension. 

About  10  minutes  after  aline  drawn  from  Alioth  to  y  Cas- 
siopeiae is  parallel  to  the  horizon,  the  pole  star  is  at  its  great- 
est distance  from  the  meridian.  As  this  is  the  case  only  once 
in  12  hours,  the  two  limits  on  the  east  side,  and  on  the  west 
side,  cannot  both  be  observed  the  same  night,  except  at  cer- 
tain seasons  of  the  year.  But  on  any  clear  night,  one  obser- 
vation may  be  made ;  and  this  is  sufficient  for  finding  a  me- 
ridian line,  if  the  distance  of  the  star  from  the  pole,  and  the 
latitude  of  the  place  be  given.  The  angle  between  the  me- 
ridian and  a  vertical  plane  passing  through  a  star,  or  an  arc 
of  the  horizon  contained  between  these  two  planes,  is  called 
the  azunuth  of  the  star.  And  by  spherioal  trigonometry, 
when  the  star  is  at  its  greatest  elongation  east  or  west, 

As  the  cosine  of  the  latitude, 
To  radius; 

So  is  the  sine  of  the  polar  distance, 
To  the  sine  of  the  azimuth. 

The  distance  of  the  pole  star  from  the  pole  in  1810,  was 
1°  42'  19".6,  and  decreases  19|^  seconds  annually. 

To  observe  the  direction  of  the  pole  star  when  its  azimuth 
is  the  greatest,  suspend  a  plumb  line  15  or  20  feet  long 
from  a  fixed  point,  with  the  weight  swinging  in  a  vessel  of  wa- 
ter, to  protect  it  from  the  action  of  the  wind.  At  the  distance 
of  12  or  15  feel  south,  fix  a  board  horizontal  on  the  top  of  a 
firm  post.  On  the  board,  place  a  sight  vane  in  such  a  man- 
ner that  it  can  slide  a  short  distance  to  the  east  or  west.  A 
little  before  the  time  when  the  star  is  at  its  greatest  elonga- 
tion, let  an  assistant  hold  a  lighted  candle  so  as  to  illuminate 
(he  plumb  line.     Then  move  the  sight  vane,  till  the  star  seen 


NOTES.  105 

through  it  is  in  the  direction  of  the  line.  Continue  to  follow 
the  motion  of  the  star,  till  it  appears  to  be  stationary  at  its 
greatest  elongation.  Then  fasten  the  sight  vane,  and  fix  a 
candle  or  some  other  object  in  the  direction  of  the  plumb 
line,  at  some  distance  beyond  it. 

As  the  declination  of  the  needle  is  continually  varying,  the 
courses  given  by  the  compass  in  old  surveys,  are  not  found  to 
agree  with  the  bearings  of  the  same  lines  at  the  present 
time.  To  prevent  the  disputes  which  arise  from  this  source, 
the  declination  should  always  be  ascertained,  and  the  courses 
stated  according  to  the  angles  which  the  lines  make  with  the 
astronomical  meridian. 

It  must  be  admitted,  after  all,  that  the  magnetic  compass 
is  but  an  imperfect  instrument.  It  is  not  used  in  the  accu- 
rate surveys  in  England.  In  the  wild  lands  in  the  United 
States,  the  lines  can  be  run  with  more  expedition  by  the 
compass,  than  in  any  other  way.  And  in  most  of  the  com- 
mon surveys,  it  answers  the  purpose  tolerably  well.  But  in 
proportion  as  the  value  of  land  is  increased,  it  becomes  im- 
portant that  the  boundaries  should  be  settled  with  precision, 
and  that  all  the  lines  should  be  referred  to  a  permanent  me- 
ridian. The  angles  of  a  field  may  be  accurately  taken  with 
a  graduated  circle  furnished  with  two  indexes.  The  bear- 
ings of  the  sides  will  then  be  given,  if  a  true  meridian  line 
be  drawn  through  any  point  of  the  perimeter. 


EXPLANATION  OF  THE  TABLES. 


Table  I  contains  the  parts  of  Mercator's  meridian,  to  eve- 
ry other  minuie.  The  parts  for  any  odd  niiniite  maybe 
found  with  sufficient  exactness,  by  takinji,  the  arithmetical 
mean  between  the  next  greater  and  the  next  less.  For  the 
uses  of  this  table,  see  Navigation,  Sec.  IIL 

Table  II  gives  the  depression  or  dip  of  the  horizon  at  sea 
for  different  heights.  Thu,^  if  the  eye  of  the  observer  is  20 
feet  above  the  level  of  the  ocean,  the  angle  of  depression  is 
4'  24".  See  Art.  99.  This  table  is  calculated  according  to 
the  rule  in  note  I,  which  gives  the  depression  59"  for  one 
foot  in  altitude ;  allowance  being  made  for  the  mean  terres- 
trial refraction. 

In  Table  III.  is  contained  the  depression  for  different 
heights  and  different  distances,  when  the  view  of  the  ocean  is 
more  or  less  obstructed  by  land.  Thus  if  the  height  of  the 
eye  is  30  feet,  and  the  distance  of  the  land  2\  miles,  the  de- 
pression is  8'.     See  Note  I. 

Table  IV  contains  the  curvature  of  the  earth,  or  the  differ- 
ence between  the  true  and  the  apparent  level  ♦or  different 
distances,  according  to  the  rule  in  art.  161.  Thus  for  a  dis- 
tance of  17  English  miles,  the  curvature  is  192  feei. 

Table  V  contains  the  distances  at  which  objects  of  differ- 
ent heights  may  be  seen  from  the  surface  of  the  ocean,  in  the 
mean  state  of  the  atmosphere.  This  is  calculated  by  first 
finding  the  distance  at  which  a  given  object  might  be  seen, 
if  there  were  no  refraction,  and  then  increasing  this  distance 
in  the  ratio  of  \/l  :  v/6.     See  Note  A. 


108  Explanation  of  the  Tables. 

Table  VI  contains  the  polar  distance,  and  the  right  ascen- 
sion in  time,  of  the  pole  star,  from  IBOO  to  1820.  From  this 
it  will  be  seen,  that  the  right  ascension  is  increasing  at  the 
rate  of  about  14  seconds  a  year,  and  that  the  north  polar  dis- 
tance is  decreasing  at  the  rate  of  19^  seconds  a  year.  From 
the  latitude  of  the  place,  and  the  polar  distance  of  the  star, 
its  azimuth  may  be  calculated,  when  it  is  at  its  greatest  dis- 
tance from  the  meridian.  The  time  when  it  passes  the  me- 
ridian may  be  ascertained  by  finding  the  difference  between 
the  right  ascension  of  th«  star  and  that  of  the  sun.  See 
Note  L, 


TABLE  I. 
MERIDIONAL  PARTS. 


M. 

0°|  l'^ 

2°  3° 

4° 

5° 

6° 

7° 

8° 

9° 

lOo 

11° 

12° 

M.; 



l_ 

1 











0 

0 

0  60 

120 

180 

240 

300 

361 

421 

482 

642  603 

664 

725 

2 

2i  62 

122 

182 

242 

302 

363 

423 

484 

544  60a 

666 

727 

2| 

4 

4'  64 

124 

184 

244 

304 

365 

425 

486 

546  607 

668 

729 

4i 

6 

6  66 

126 

186 

246 

306 

367 

427 

488 

5481609 

670 

731 

6 

8 

8  68 

128 

188 

248 

308 

369 

429 

490 

550 

611 

672 

734 

8 

10 

10  70 

130 

190 

250 

310 

371 

431 

492 

552 

613 

674 

736 

10 

12 

12  72 

132 

192 

252 

312 

373 

433 

494 

554 

615 

676 

738 

12 

14 

14  74 

134 

194 

254 

314 

375 

435 

496 

556 

617 

678 

740 

14 

16 

16  76 

136 

196 

256 

316 

377 

437 

498 

558 

619 

680 

742 

16 

18 

18  78 

138 

198 

258 

318 

379 

139 

500 

560 

621 

682 

744 

18 

20 

20  80 

140 

200 

260 

320 

381 

441 

502 

562 

623 

684 

746 

20 

22 

22  82 

142 

202 

262 

322 

383 

443 

504 

565 

625 

687 

748 

22 

24 

24  84 

144 

204 

264 

324 

385 

445 

506 

567 

627 

689 

750 

24 

26 

26 

86 

146 

206 

266 

326 

387 

447 

508 

569 

629 

691 

752 

26 

28 

28 

88 

148 

208 

268 

328 

389 

449 

510 

571 

632 

693 

754 

28 

30 

30 

90 

150 

210 

270 

331 

391 

451 

512 

573 

634 

695 

756 

30 

32 

32 

92 

152 

212 

272 

333 

393 

453 

514 

575 

636 

697 

758 

32 

34 

34 

94 

154 

214 

274 

335 

395 

455 

516 

577 

638 

699 

76034 

36 

36 

96 

156 

216 

276 

337 

397 

457 

518 

579 

640 

701 

762  36 

38 

38 

98 

158 

218 

278 

339 

399 

459 

520 

581 

642 

703 

764 

381 

40 

40 

100 

160 

220 

280 

341 

401 

461 

522 

583 

644 

705 

766 

40 

42 

42 

102 

162 

222 

282 

343 

403 

463 

524 

585 

646 

707 

768;42 

44 

44 

104 

164 

224 

284 

345 

405 

465 

526 

587 

648 

709 

770  44 

46 

46 

106 

166 

226 

286 

347 

407 

467 

528 

589 

650 

711 

772 

46 

48 

48 

108 

168 

228 

288 

349 

409 

469 

530 

591 

652 

713 

774 

48 

50 

50 

110 

170 

230 

290 

351 

411 

471 

532 

593 

654 

715 

777 

'50 

52 

52 

112 

172 

232 

292 

353 

413 

473 

534 

595 

656 

717 

779 

52 

54 

54 

114 

174 

234 

294 

355 

415 

476 

536 

597  658 

719 

781 

;54 

56 

56 

116 

176 

236 

296 

357 

417 

478 

538 

599  660 

721 

783 

,56 

58 

68 

118 

178 

238 

298 

359 

419 

480 

540 

601 

662 

723 

785 

:58 





















1 

M. 

0° 

r 

20 

3° 

4° 

5° 

6° 

r 

8° 

9°|lO° 

11° 

12° 

|M. 

TABLE  I. 
MERIDIONAL  PARTS. 


M. 

13° 

14° 

150  16° 

17°  18°  j  19° 

200 

21° 

22° 

VI. 

(; 

787 

848 

010,  973 

1036|1098 

- 
1161 

1225 

1289 

1354 

0 

2 

789 

85i 

(»13,  975 

I037|1I00 

1164 

1227 

1291 

1366 

2 

4 

791 

bo.s 

9!..'> 

977 

1039 

1102 

1166 

1229  1293 

1358 

4 

c 

793 

'cbo 

917 

979 

1042 

1105 

1168 

1232 

1296 

1360 

6 

8 

79:-> 

857 

919 

981 

1044 

1107 

1170 

1234 

1298 

1362 

« 

10 

797 

85'' 

921 

983J1046 

1109 

1172 

1236 

1300 

1364 

10 

12 

799 

661 

923 

985  1048 

nil 

1174 

1238 1302 

1367 

12 

14]801 

8bo 

'32 

987J1050 

1113 

1176 

124011304 

1369 

14 

16803 

;)65 

927 

9891 1052 

1115 

178 

I242il306 

1371 

16 

18S05 

86-; 

929 

991 

1054 

1117 

1181 

1244 

1308 

1373 

18 

20  807 

SG9 

031 

994 

1056 

1119 

1183 

1246 

1311 

1376 

20 

22809 

871 

933 

996|l058!l]21 

1185 

1249 

1313 

1377 

0.2 

24:811 

87,i 

935 

998!1000!ll23 

1187 

1251 

1316 

1380 

24 

26  813 

87j 

937;  1000;  1063 

1126 

1189 

1263 

1317 

1382 

26 

28  816 

877 

939  1002 

1065 

-1128 

1191 

1265 

1319 

1384 

28 

30;818 

87P 

942  1004 

1067 

1130 

1193 

1267 

1321 

1386 

30 

321820 

882 

944  1006 

1009 

1132 

1195 

1269!  1324 

1388 

32 

34 

822 

>i^-l 

946;  1008 

1071 

113411198 

1261  1326 

1 390 

34 

36 

824 

8.6 

948il010 

1073 

113e5 

1200 

1264  1328 

1393 

36 

38 

826 

88Ji 

950 

1012 

1075 

1138 

1202 

1266 

13.'.0 

1395 

38 

40 

828 

890 

952 

1014 

1077 

1140 

1204 

1268 

1332 

1397 

40 

42830 

892 

9rA 

1016 

1079 

11 42' 1206 

1270 

1334 

1399 

42 

441832 

894 

966 

1019 

IO81 

1145 

1208 

1272 

1 336 

1401 

44 

46834 

896 

968 

1021 

10S4 

1147 

1210 

1274 

1339 

1403 

46 

48  836 

898 

960 

1023 

1086 

1149 

1212 

1276 

1341 

1406 

48 

60|838 

f'OO 

962 

1025 

1088 

1151  1215 

1278 

i343 

1408 

50 

52|840 

9U2 

964 

1027 

1 090 

1163 

]2|7 

1281 

1346 

1410 

52 

54 

842 

901 

966;  10291 1092 

1166 

1219 

128.: 

1347 

1412 

64 

56 

844 

906  969;  1 031  i  1094 

1167 

1221 

1286 

1349 

1414 

56 

t 

846 

908 

971 1 1033 

1096 

1159 

1223 

1287 

1362 

1416 

22° 

58 

1 

L 
JM. 

13° 

140 

15°|  16° 

17° 

18° 

1 9° 

20° 

TABLE  I. 
MERIDIONAL  PARTS, 


0 

23° 

24° 

25° 

26°  27° 

1  28°  1  29°  1  30° 

31° 

32° 

31. 
0 

|l419 

1484 

1550 

1616 

1684 

jl751  1819  1888  1950 

2028 

2 

Il421 

1486 

1552 

1619 

16.6 

11753  1822|'i8'^*l  1960 

2031 

2 

4 

14L'3 

1488 

i554 

1621 

1688 

1756 

1824 

1893J1963 

2033 

4 

6 

1423 

1491 

1557 

1623 

1690 

1758 

1826 

1895 

1965 

2035 

6 

8 

1427 

1493 

1559 

1625 

1693 

1760 

1829 

1898 

1967 

2038 

8 

10 

1430 

1495 

1561 

1628 

1695 

1762 

1831 

1900 

1970 

2040 

10 

1 '' 

143i 

1497 

1563 

163{j 

1697 

1765 

1833 

1902 

1972 

2043 

12 

14 

1434 

1499 

1565 

I G3- 

1699 

1767 

1835 

1905 

1974 

2045 

14 

16 

1435 

150211568 

1634 

J701 

1769 

1838 

1907 

1977 

2047 

]e 

IG 

1438 

1504 

1570 

1637 

1704 

1772 

1840 

1909 

1979 

2050 

18 

20 

1440 

1506 

1572 

1639 

1706 

1774 

1842 

1912 

1981 

2052120 

22 

1443 

1 508 

1574 

1''41 

1  ro8 

1776 

1845 

1914  1984 

2054122 

24 

1415 

1510 

1577 

1643 

1711 

1778 

1847 

1916 

1986 

2057:24 

26 

1447 

151.S  1579 

1645 

1713 

1781 

1849 

1918 

1988 

2059126 

28 

1449 

1515 

1581 

1648 

1715 

1783 

1852 

1921 

1991 

2061  !28 

30 

1451 

1517il583 

1650 

1717 

1785 

1854 

1923 

1993 

2064  30 

3-> 

1453 

1519  \oSo 

iu5^ 

1720 

1787 

1856 

1925 

1995 

2066  32 

34 

1456 

lo2i 

I588|i654l 

172J 

1790 

1858 

1928 

1998 

2069  34 

[30 

145" 

152  ^ 

1590 

1657 

1724 

*1792 

1861 

1930 

2000 

2071  ,'36 

36 

1460 

1526 

1592 

1659 

1726 

1794  1863 

1932 

2002 

2073 

38 

40 

1462 

1528 

1594 

1661 

1729 

1797jl865  1935 

2005 

2076 

40 

42 

14^4 

•530 

1596 

1663 

1731 

1799  1868  1937 

2007 

2078 

42 

44 

1467 

1532 

15i?9 

1666 

1733 

1801  1870  1939 

2010 

2080 

44 

4d 

1469 

1535 

J601 

1668 

1735 

1:03  1872  1942 

2012 

2083 

46 

48 

1471 

1637 

1603 

1670 

1738 

1806 

1875  1944 

2014 

2085 

48 

50 

1473 

1539|1605 

1672 

1740 

1808 

1877  1946 

2017 

2088 

50 

52 

1475 

154i|1608 

1675 

1742 

1810 

1879 

1949 

2019 

2090 

52 

54 

1477 

1513I1610 

1677  1 7441 

1813 

1881 

1951 

2021 

2092 

54 

56 

1480  1546, 1612'!  t>79j  17471 1815!  1884, 1953 

2024 

2095 

d6 

58 

1482  154e  161411681  1749| 

1817 
28° 

1836  1956 

2026 

2097 

58 

M. 

23°  24°  1  25°  26° 

27°  1 

29°  30° 

31° 

32°  >!.! 

TABLE  1. 
MERIDIONAL  PARTS. 


0 

33° 

34" 

36" 

36" 

37° 

38° 

39°  \  40° 

f 

41" 

42° 

JVl. 

2100 

2171 

2244 

2318 

2393 

2468 

2546 

2623 

2702 

2782 

0 

2 

2102 

2174 

2247 

2320 

2396 

2471 

2648 

2625 

2704 

2784 

2 

4 

2104 

217C 

2249 

2323 

2398 

2473 

2660 

2628 

2707 

2787 

4 

6 

2107 

2179 

2262 

2326 

2400 

2476 

26631 263  ij  27  10 

2790 

6 

8 

2109 

2181 

2264 

2328 

2403 

2478 

2665 

2633 

2712 

2792 

8 

10 

2111 

2184 

2251 

2330 

2405 

2481 

2568 

2636 

2716 

2795 

10 

12 

2114 

2186 

2269 

2333 

2408 

2484 

2560 

2638 

2718 

2798 

12 

142116 

2188 

2261 

2336 

2410 

2486 

2663 

264  i 

2720 

2801 

14 

16 

2119 

2191 

2261 

2338 

2413 

2489 

2566 

2644 

2723 

2803 

16 

18  2121 

2193 

2266 

2340 

2415 

249J 

2668 

2646 

2726 

2806 

18 

20  2123 

2196 

2269 

2343 

2418 

2494 

2571 

2649 

2728 

2809 

20 

22  2126 

2198 

2271 

2345 

2420 

2496 

2573 

2661 

2731 

2811 

22 

24 

2128 

2200 

2274 

2348 

2423 

2499 

267c  (2654 

2733 

2814 

24 

26 

2131 

2203 

2276 

2350 

2425 

2501 

2678 

2657 

2736 

2817 

26| 

28 

2133 

2206 

2279 

2353 

2428 

2504 

2581 

2669 

2739 

2820 

28 

30 

2136 

2208 

2281 

2355 

2430 

2506 

2584 

2662 

2742 

2822 

30 

32 

2138 

2210 

2283 

2358 

2433 

2509 

2586!2665 

2744 

2826 

32 

34 

2140 

2213 

2286 

2360  2435 

251212589 

2667 

2747 

2828,34i 

36 

2143 

2216 

2288 

2363  2438 

2514|2591 

2670 

2760 

2830 

36 

38 

2145 

2217 

2291 

2365 

2440 

2617 

2694 

2673 

2762 

2833 

38 

40 

2147 

2220 

2293 

2368 

2443 

2619 

2697 

2675 

2755 

2836 

40 

42 

2150 

2222 

2296 

2370 

2445 

2622 

2599 

2678 

2768 

2839 

42 

44 

2162 

2226 

2298 

2373 

2448 

2524 

2602 

2680 

2760 

2841 

44 

46 

2165 

2227 

2301 

2375 

2461 

2627 

2604 

2683 

2763 

2844 

46 

48 

2167 

2230 

2303 

2378 

2453 

2630 

2607 

2686 

2766 

2847 

48 

50 

2169 

2232 

2306 

2380 

2468 

2532 

2610 

2688 

2768 

2849 

50 

62 

2162 

2236 

2308 

2383  2458 

2535 

2612 

2691 

2771 

2862 

62 

64 

2164 

2237 

2311 

2385 

2461 

2537 

2616 

2694 

2774 

2865 

64 

6612167 

22.39 

2313 

2388 

2463 

2540 

2617 

2696 

2776 

2858 

56 

68  2169 

) 

2242 

2316 

2390 

2466 

2542  2620 

2699  2779 

2860 

68 

JM. 

33° 

34» 

36° 

36° 

37° 

38°  39° 

40°  41° 

42° 

M. 

TABLE  1. 
MERIDIONAL  PARTS. 


10 
12 
14 
16 

18 

20 
22 
24 
26 
28 

30 
32 
34 
'36 
38 

40 
42 
44 
46 


43°  44°  45°  46°  4?°  48°  49 


2863 
2866 
2869 
2871 
2874 

2877 
2880 
2882 
2885 
2888 

2891 
2893 
2896 
2899 


2946 
2949 
2951 
2954 
2957 


3030 
3033 
3036 
3038 
3041 


2960  3044 
2963  3047 


2902  2985 


2904 
2907 
2910 
2913 
2915 

2918 
2921 
2924 
2926 
48  2929 


50 
52 
54 
56 

58 


2965 
2968 
2971 

2974 
2976 
2979 
2982 


2988 
2991 
2993 
2996 
2999 

3002 
3005 
3007 
3010 
3013 


2932  3016  3101 


3050 
3053 
3055 

3058 
3061 
3064 
3067 
3070 

3073 
3075 
3078 
3081 
3084 


3116 
3118 
3121 
3124 
3127 

3130 
3133 
3136 
3139 
3142 

3144 
3147 
3150 
3153 
3156 

3159 
3162 
3165 
3168 
3171 


3087  3173 
3090  3176 
30933179 


3095 
3098 


2935|3019 


2937 
2940 
2943 


M.  43° 


3021 
3024 
3027 


3104 
3107 
3110 
3113 


3182 
3185 

3188 
3191 
3194 
3197 
3200 


44°  45°  460  47° 


3203 
3206 
3209 
3212 
3214 

3217 
3220 
3223 
3226 
3229 

3232 
3235 
3238 
3241 
3244 

3247 
3250 
3253 
3256 
3259 

3262 
3265 
3268 
3271 
3274 

327r 
3280 
3283 
3286 
3289 


3292 
3295 
3298 
3301 
3303 


3382 
3385 
3388 


50^ 


3474 
3478 
3481 


3391  3484 

3394  3487 


3306  3397 
3309  3400 


3312 
3316 
3319 

3322 
3325 
3328 
3331 
3334 

3337 
3340 
3343 
3346 
3349 

3352 
3355 
3358 
3361 
3364 

3367 
3370 
3373 
3376 
3379 


48^ 


3403 
3407 
3410 

3413 
3416 
3419 
3422 
3425 

3428 
3431 
3434 
3437 
3440 

3443 
3447 
3450 
3453 
3456 


5r 


3569 
3572 
3575 
3578 
3582 

3585 
3588 
3591 


3490 

3493 

3496 

349913594 

350313598 

350GI3601 


3509 
3512 
3515 
3518 

3521 
3525 
3528 
3531 
3534 

3537 
3540 
3543 
3547 
3550 


3459  3553 


3462 
3465 
3468 
3471 


49^ 


3556 
3559 
3562 
3566 


50°  51 


3604 
3607 
3610 
3614 

3617 
3620 
3623 
3626 
3630 

3633 
3636 
3639 
3643 
3646 

3649 
3652 
3655 
3659 
3662 


52°  M. 


3665 
3668 
3672 
3675 
3678 

3681 
3685 
3688 
3691 


J695  IS 


3698 
3701 
3704 
3708 
3711 

3714 
3717 
3721 
3724 
3727 

3731 
3734 
3737 
3741 
3744 

3747 
3750 
3754 
3757 
3760 


20 

22 
24 
26 

28 

30 
32 
34 
36 

38 

40 

42 
44 
46 

48 

50 
52 
54 
56 

58 


52° 


16 


TABLE  I. 
MERIDIONAL  PARTS. 


>1.  53= 


54^ 


55" 


56^ 


57^ 


5&0 


59= 


60' 


61 


62" 


10 
12 
14 
16 

18 

120 
|22 
24 
i26 


3764 
3767 
3770 
3774 
3777 

3780 
3784 
3787 
3790 
3794 

3797 
3800 
3804 
3807 


386o 
3868 
3871 
3875 
3878 

3882 
3885 
3889 
3892 
3895 

3899 
3902 
3906 
3909 


3968 
3971 
3975 
3978 
3982 

3985 
3989 
3992 
3996 


4074 
4077 
4081 
4085 
4088 

4092 
4095 
4099 
4103 


3999  4106 


4003 
4006 
4010 
4014 


3811  3913 


30 

32 
34 
36 
38 

40 
42 
44 

46 
48 

I 


3814 
3817 
3821 
3824 
3827 

3831 
3834 
3838 
3841 
3844 

3848 
3851 
3854 
3858 


4110 
4113 
4117 
4121 


4017|4124 


50 
52 
54 
56 
58  3861 


3916 
3919 
3923 
3926 
3930 


3933 
3937 
3940 
3944 
3947 

3951 
3954 
3958 
3961 
3964 


4021 
4024 
4028 
4031 


4128 
4132 
4135 
4139 


4183 
4186 
4190 
4194 
4197 

4201 
4205 
4208 
4212 
4216 

4220 
4223 
4227 
4231 
4234 

4238 
4242 


4294 
4298 
4302 
4306 
4309 

4313 
4317 
4321 
4325 


4409 
4413 
4417 
4421 
4425 

4429 
4433 
4436 
4440 


4328  4444 


4332 
4336 
4340 
4344 
4347 

4351 
4355 


4448 
4452 
4456 
4460 
4464 

4468 
4472 
4476 
4480 


4035  4142 


4038 
4042 
4045 
4049 
4052 

4056 
4060 
4063 


4146 
4150 
4153 
4157 
4161 

4164 
4168 
4172 


424614359 
424914363 
4253  436714484 


4257 
4260 
4264 
4268 
4272 


4067  4175 
4070  4179 


4370  4488 


4374 
4378 
4382 
4386 


4275  4390 


4279 
4283 
4287 
4291 


4394 
4398 
4401 
4405 


4527 
4531 
4535 
4539 
4543 

4547 
4551 
4555 
4559 
4564 

4568 
4572 
4576 
4580 
4584 

4588 
4592 
4596 
4600 
4604 


4608 
4492  4612 


4495 
4499 
4503 

4507 
4511 
4515 
4519 
4523 


4616 
4620 
4625 

4629 
4633 
4637 
4641 
4645 


4649 
4653 
4657 
4662 
4666 

4670 
4674 
4678 
4682 
4687 

4691 
4695 
4699 
4703 
4707 

4712 
4716 
4720 
4724 
4728 

4733 
4737 
4741 
4745 
4750 

4754 
4758 
4762 
4766 
4771 


4775 
4779 
4784 
4788 
4792 

4796 
4801 
4805 
4809 
4814 

4818 
4822 
4826 
4831 
4835 

4839 
4844 
4848 
4852 
4857 

4861 
4865 
4870 
4874 
4879 

4883 
4887 
4892 
4896 


4901  58 


0 
2 
4 
6 
8 

10 
12 
14 

16 
18 

20 
22 
24 
26 
28 

30 
32 
34 
36 
38 

40 
42 
44 
46 
48 

50 
52 
54 
56 


53° 


54« 


55° 


56°  57° 


58°|59*>  60o|61<»  62°  M 


TABLE  r 
MERIDIONAL  PARTS. 


M  63°  64°  65 


10 
12 
14 

16 

1 

20 

22 
24 
f26 

28 


3014972 


32 
34 
36 
38 


4905 
4909 
4914 
4918 
4923 

4927 
4931 
4936 
4940 
4945 

4949 
4934 
4958 
4963 
4967 


4976 
4981 
4985 
4990 


40  4994 
42  4999 
5003 
5008 
5012 


5017 
5021 
5026 
5030 
5035 


5039 
5044 
5049 
5053 
5058 

5062 
5067 
5071 
5076 
5081 

5085 
5090 
5095 
5099 
5104 

5108 
5113 
5118 
5122 
5127 

5132 
5136 
5141 
5146 
5151 

5155 
5160 
5165 
5/69 
5174 


5179 
5184 
5188 
5193 
5198 

5203 
5207 
5212 
5217 


66= 


67°    680  69°    70°  71°  72° 


5324  5474 
53285479 
5333  5484 


5338 
5343 

5348 
5353 
5358 
5363 


5222  5368 


5226 
5231 
5236 
5241 
5246 

5250 
5255 
5260 
5265 
5270 

5275 
5280 
5284 
5289 
5294 

5299 
5304 
5309 
5314 
5319 


M.   63°     64°    65°     66°     67° 


5489 
5495 

5500 
5505 
5510 
5515 


5631 
5636 


5795  5966 
5800  5972 
5642]5806|5978 
5647  581 1|5984 
565215817  5989 


5658 
5663 
5668 
5674 
55205679 


5373 
5378 
5383 
5388 
5393 

5398 
5403 
5408 
5413 
5418 

5423 
j428 
5433 
5438 
5443 

5448 
5454 
5459 
5464 
5469 


5526 
5531 
5536 


5685 
5690 
5695 


5541  5701 
5546  5706 

5552  5712 
55575717 
5562I5723 
5567:5728 
557315734 

5573  5739 
5583  5745 
5588(5750 
5594  5756 
5599  5761 


5823 
5828 
5834 
5839 


5995 
6001 
6OO7 
6013 


5845  6019 


5851 
5856 
5862 
5868 
:;874 


6025 
6031 
6037 
6043 
6049 


587916055 


5604 
5610 
5615 
5620 
5625 


5767 
5772 
5778 


5885 
5891 
5896 
5902 

59O8 
5914 
5919 
5925 
5931 

5937 
5943 
5948 


5783  5954 
5789  5960 


6061 
6067 
6O73 


6146 
6152 
6158 
6164 
6 170 

6177 
6183 


6335 
6341 
6348 
6354 
6361 

6367 
6374 


618916380 
6195  6387 
6201  6394 


6208 
6214 
6220 
6226 
623vi 

6239 
6245 
6252 
6258 


6079  6264 


6085 
6091 
6097 
6103 
6109 

6115 
6121 
6127 
6133 
6140 


6400 
6407 
6413 
6420 

6427 

6433 
6440 
6447 
6453 
6460 


6271  6467 
6277  6473 


680  690  jQo    7-1  o  j2 


6283 
6290 
6296 

6303 
6309 
6315 
6322 
6328 


6480 
6487 
6494 

6500 
6507 
9614 
6521 
6528 


0 

2 
4 
6 

8 

10 
12 
14 
16 

18 

20 

22 
24 
26 
28 

30 

32 
34 
36 
38 

40 
42 
44 
46 
48 

50 

52 
54 
56 

58 

• 


TABLE  I. 
MERIDIONAL   PARTS. 


M.  73°  74^^  75^'  76"  I  77°  78 


0  6534 


6541 
6548 
6555 
6562 

6569 
6576 
6583 
6590 
6597 

6603 
6610 

6617 
6624 
6631 

6639 
6646 
6653 
6660 
6667 

6674 
6681 
6688 
6695 
6702 

6710 
6717 
6724 
6731 
6738 


6746 
6753 
6760 
6768 
6775 

6782 
6790 
6797 
6804 
6812 

6819 
6826 
6834 
6841 
6849 


6970 
6978 
6986 
6994 
7001 

7009 
7017 
7025 
7033 
7041 

7048 
7056 
7064 
7072 


7210 
7218 

7227 
7235 


7467 
7476 
7485 
7494 


7243J7503 

7252|7512 
72607521 
72687530 


7277 
7285 

7294 
7302 
7311 
7319 


7080  7328 


6850  7088 
6864j7096 
6871i7104 
6879;7112 
6886  7120 


6894 
6901 
6909 
6917 
6924 


7128 
7136 
7145 
7153 
7161 


7336 
7345 
7353 
7362 
7371 

7379 
7388 
7397 
7406 
7414 


6932  7169  7423 

6939|7177 

69477185 


6955 
6963 


M.|  73°  74°  75- 


7194 

7202 


7432 
7441 
7449 
7458 


76' 


7539 
7548 

7557 
7566 
7576 
7585 
7594 

7603 
7612 
7622 
7631 
7640 


7745 
7754 

7764 
7774 
7783 


7793 
7803 
7813 
7822 
7832 


79^ 


8046 
8056 
8067 
8077, 


80-     81°  82"  M. 


8375  8739 
8387  8752 


8398 
8410 
8088;8422 

I 


8099 
8109 
8120 
8131 
8141 


7842  8152  8492 


8433 
8445 
8457 
8469 
8480 


7852 
7862 
7872 
7882 


81638504 
817418516 
8185  8528 
8196  8540 


7892|8207 
79028218 
791218229 
7922|8240 
7932|8251 


76507942 
7659,7953 
76687963 
76787973 
76877983 


76977994 
770618004 
7716^8014 


7725 
7735 


8025 


8765 
8778 
8791 

8804 
8817 
8830 
8843 
8856 

8869 
8883 
8896 
8909 
8923 

8936 
8950 
8963 
8977 
8991 

9005 
9018 
9032 
9046 
9060 


8318  8676  9074 
832918688  9088 


8552 
8565 
8577 
8589 
8601 


8262  8614 


8273 
8284 


8626 
8638 


8295;8651 
830718663 


9145 
9160 
9174 
9189 
9203 

921 

9233 

9248 

9262 

9277 

9292 
9307 
9322 
9337 
9353 

9368 
9383 
9399 
9414 
9430  38 

9445  40 


10 
12 
14J 
16 
18 

20 

22] 
24 
26 
28 

30 
32 
34 
36 


9461 
9477 
9493 
9509 


42 
44 

4Q\ 
48 


9341 
8352 


8035  8364 


8701 
8714 
8726 


77°  78°  79°  80°  81°  82"  M. 


9525  50 
9541  52 
54 


91039557 
91179573 
913l|9589 


TABLE  II. 


TABLE  HI. 


Depression  of  the  Ho- 
rizon of  the  Sea. 


Dip  oi  the  Sea,   at  different   Distances   from 
the  Observer. 


Depres- 
sion. 


0*  39" 


1 

24 

1 

42 

1 

58 

2 

12 

2 

25 

3  36 

2  47 

2 

57 

3 

7 

3 

16 

3  25 

3  33 

3 

41 

3  48 

3 

56 

4 

3 

4 

10 

4 

17 

4 

24 

4 

37 

4 

49 

5 

1 

5 

13 

5 

23 

5 

49 

6 

14 

6  36 
6  57 


37 
14 


8  48 

9  20 
9  50 

10  47 

11  39 

12  27 

13  12 
13  55 


Dist.  ot 
land  m 

Height  of    the   Eye  above  the   &ea, 
in  (eet. 

mile/. 

5  |I0  |15  |20  |25  |30  |35  |40 

0  i 

11' 

22'  34'145';56"6B'|79';90} 

0  i 

6 

11 

17  122  128  ,34 

39 

45 

0  a 

4 

8 

12 

15   19  23 

27 

30 

1  0 

4 

6 

9 

12   15   17 

20 

23 

I  i 

3 

5 

7 

9 

12  114 

16 

19 

1  ; 

3 

4 

6 

S 

10 

11 

14   15 

2  0 

2 

3 

5 

6 

8 

lO 

11    12 

2  ^ 

2 

3 

5 

6 

7 

8 

9 

10 

3  0 

2 

3 

4 

5 

6 

7 

8 

8 

3  i 

2 

3 

4 

5 

6 

6 

7 

7 

4  0 

2 

3 

4 

4 

5 

6 

7 

7 

5  0 

2 

3 

4 

4 

5 

5 

6 

6 

6  0 

2 

3 

4 

4 

5 

5 

6 

6 

TABLE  IV, 

Curvature  of  the  Earth. 


Dist.  in 

tJ  eight. 

Dist.  id 

Height. 

;i 

1 

[oches. 

Feet. 

1 
4 

1 

2 

15 

149 

i 

2 

16 

170 

1 

8 

i          17 

192 

Feet. 

!         18 

215 

2 

2.6 

i         ^^ 

240 

3 

6. 

!         20 

266 

4 

10.6 

'         25 

4i5 

5 

16.6 

30 

599 

6 

23.9     j 

35 

814 

7 

32.5     : 

40 

1064 

8 

42.5     : 

45 

1346 

9 

53.8     j 

66.4     I 

50 

1662 

10 

60 

2394 

11 

80.2     1 

70 

3258 

12 

95.4     1 

80 

4255 

13 

112. 

90 

5386 

14 

130 

100 

6649 

TABLE  V. 

Distances  at  which  Objects  can  be 
seen  at  Sea. 


Height  ir, 

Distance    in 

Height  ii 

)    Distance    in 

leet. 

Eng.  miles. 

feet. 

Eng.  miles, 

1 

1.3 

60 

10.2 

2 

1.9 

70 

11.1 

3 

2.3 

80 

11.8 

4 

2.6 

90 

12.5 

5 

2.9 

100 

132 

6 

32 

200 

18.7 

7 

3.5 

300 

22.9 

8 

3.7 

400 

26.5 

9 

4. 

500 

29.6 

10 

42 

600 

32.4 

12 

4.6 

700 

35. 

14 

49 

800 

374 

16 

53 

900 

39.7 

18 

5.6 

1000 

41.8 

20 

59 

2000 

592 

25 

66 

3000 

72-5 

30 

7.3 

4000 

83-7 

35 

7.8 

5000 

93-5 

40 

8.4 

10000 

133- 

45 

8.0 

15000 

163- 

50 

9.4 

20000 

188- 

TABLE  VI. 

The  Polar  Distance  and  Right  Ascension  af  the 
Pole  Star. 


Polar   Distance. 


1800 

io 

45'  35" 

1801 

45  15 

1802 

44  56 

1803 

44  36 

1804 

44  17 

1S05 

43  57 

1806 

1 

43  33 

1807 

43  18 

1808 

4i   58 

1809 

42  39 

1810 

42  19 

1811 

42 

18l£ 

41  40 

1813 

41  21 

1814 

41   1 

18)5 

40  42 

1816 

40  23 

1817 

40  04 

1818 

39  45 

1819 

39  25 

1820 

39  05 

Ann.  Var. 


--19".5 


I— 19".4 


Right  Ascension.  | 

Ann.  Var.  \ 

h.   m. 

8. 

s. 

0  52 

24 

+  12.9 

52 

37 

52 

50 

53 

3 

53 

16 

53 

20 

53 

42 

53 

55 

54 

9 

54 

22 

s. 

54 

m 

4-136 

54 

50 

55 

04 

55 

18 

55 

33 

55 

47 

56 

02 

56 

17 

56 

32 

56 

46 

s. 

57 

01 

+  14.3 

S2=«ilj£RVErijrG 


yv.y/ 


« 

A^ 

^ 

-^C' 

5 
» 

^rt 

B 

c 


.3^ 


V.^.iT.j.^^^^^   .,.^ 


^'m 


N'  and  SV^RVBYIXG.  /^m. 


^        B 


JJ       _D 


JTAVIBATIOX  and  SrBVBYilVC.  f^m 


AN 


ELEMENTARY 

TREATISE 


CONIC  SECTIONS,  SPHERICAL  GEOMETRY. 


AJfS 


SPHERICAL  TRIGONOMETRY. 


BY  MATTHEW  R.  BUTTON, 

PROFESSOR   01   MATHEMATICS   AND   NATURAL   PHIL080PHT 
IN    YALE    COLLEGE. 


BEING  THE  FIFTH  AND  SIXTH  PARTS 

0»  A   COURSE   OF    MATHEMATICS   ADAPTED   TO    THE   METHOD  OE 
INSTRUCTION    IN    AMERICAN   COLLEGES 


NEW-HAVEN. 

PUBLISHED  BY  HOWE  &  SPALDING. 

S.  Converse,  printer. 

1824. 


DISTRICT  OF  CONNECTICUT,  ss. 

BE  IT  REMEMBERED,  That  00  the  seventeenth  day  of  March,  in 
the  forty-ei§:th  year  of  the  InJependence  of  th     United  States  of 
America,  Matthew  R.  Dutton,  of  the  said  District,  hath  de- 
posited in  this  Otfice  the  title  of  a  book,  the  right  whereof  he  clainjfc 
as  Author,  in  the  words  following,  to  wit :  — 
•'  An  Elementary  Treatise  on  Conic  Sections,  Spherical  Geometry,  and  Spher- 
<*ical  Trigonometry.     By  Matthew  R.  Dutton,  Profnssor  of  Mathematics  and  Nat- 
♦*  ural  Philosophy  in  Yale  College.     Being  the  Fifth  and  Sixth  Part?  of  a  course  of 
"Mathematics  adapted  to  the  method  of  instruction  in  American  Colleges." 

In  conformity  to  the  act  of  the  Congress  of  the  United  States,  entitled,  "  An  Act 
for  the  encouragement  of  learning,  by  securing  the  copies  of  Maps,  Charts  and 
Books,  to  the  authors  and  proprietors  of  such  copies,  during  the  times  therein  men- 
tioned." 

CHAS.  A.  INGERSOLL, 
Clerk  of  the  District  of  Connecticut. 
A  true  copy  of  Record,  examined  and  sealed  by  me, 

CHAS.  A.  INGERSOLL, 
Clerk  of  the  District  of  Connecticut, 


ADVERTISEMENT. 

In  preparing  the  following  Treatise  on  Conic  Sections,  it  was 
ietermined  to  adopt  the  general  plan  and  arrangement  of  propo- 
iitions  in  Button's  Conic  Sections  ;  the  deviations  from  that 
plan  however  have  become  so  numerous  and  considerable  as  to 
leave  but  a  slight  resemblance  to  it.  The  authors  whose  works 
tiave  been  consulted,  and  in  some  cases  freely  used,  are  Appollo- 
vius,  Archimedes,  Hamilton,  Simpson,  West,  Vince,  Emerson 
and  Bridge  ;  especially  the  two  authors  last  mentioned.  This 
general  acknowledgment  is  made,  as  it  is  inconvenient  or  im- 
possible to  specify  more  particularly  what  is  due  to  each. 

The  authors  whose  works  have  supplied  the  principal  materi- 
als for  the  Spherical  Geometry  and  Trigonometry,  are  Theodo- 
sius,  Simpson,  Platfair,  and  Cagnoli  ;  the  Projections  and 
Examples  are  taken  with  slight  alterations  from  the  Mathematics 
of  Webuer,  for,  in  the  following  work,  there  has  been  no  af- 
fectation of  originality,  when  there  appeared  no  room  for  improve- 
ment. ^  M.  R.  D. 


CONTENTS, 


CONIC  SECTIONS. 

Definitions,         ------ 

Parabola,        ------. 

Ellipse,        ------- 

HyPERBOLA,       ------- 

Curvature  of  Conic  Sections,  -         -         - 
APPENDIX  TO  CONIC  SECTIONS, 

1.  Similar  and  Sub-Contrary  Sections, 

2.  Comparison  of  Conic  Sections,     - 

SPHERICAL  GEOMETRY. 

Definitions  and  Sections  of  the  Sphere, 
Spherical  Triangles,    -         -         -         -         - 

IWTErtSECTIONS  <5lz:C.         -  -  -  -  - 

APPENDIX  TO  SPHERICAL  GEOMETRY, 

Projection,  ----- 

Problems  in  Stereographic  Projection, 

SPHERICAL  TRIGONOMETRY. 

Geometrical  Principles  of         -         -         - 
Arithmetical  Calculation  „         -         -         - 
Notes,        .-.--,• 


CONIC  SECTIONS. 


DEFINITIONS. 

1.  CONIC  SECTIONS  are  the  figures  made  by  the 
mutual  intersection  of  a  cone  and  plane. 

2.  According  to  the  different  positions  of  the  plane,  five 
different  figures  or  sections,  are  produced  ;  viz,  a  Triangle^ 
a  Circle^  a  Parabola^  an  Ellipse^  and  a  Hyperbola. 

3.  If  the  plane  pass  through  the  vertex  of  the  cone, 
and  any  part  of  the  base,  the  section  will  be  a  triangle. 

Case  1.  Let  it  pass  through  the  vertex,  perpendicular 
to  the  base,  and  coinciding  with  the  axis  of  the  cone;  it 
will  then  coincide  with  the  triangle  by  the  revolution  of 
which  the  cone  is  generated.     The  section  therefore  willgup.s.ii. 
be  an  isoceles  triangle  VCD,  whose  angle  at  the  vertex  Fig.  i 
will  be  double  that  of  the  generating  triangle. 

Case  2.  Let  the  plane  now  cut  the  base  obliquely. 
This  section  also  will  be  a  triangle,  for  every  straight 
line  drawn  from  the  vertex  of  a  cone  to  any  point  of 
the  circumference  of  its  base,  is  in  the  surface  of  the 
cone ;  since  it  coincides  with  the  Hypothenuse  of  the 
revolving  triangle  b)  which  the  cone  is  generated. 

4.  If  the  plane  cut  the  cone  parallel  to  its  base,  the  sec-  Fig;.  2. 
tioH  will  be  a  circle,  as  ABD  — for  this  section  wll  coin- 
cide with  the  revolution  of  a  line  perpendicular  to  the  axis 

of  the   generating  triangle,  and  will  therefore  be  a  circle, 
whose  radius  is  equal  to  that  perpendicular. 

5.  If  the  cone  be  cut  by  a  plane  parallel  to  one  of  its 
sides,  the  section  is  called  a  Pt/ra6o/a;    as  ACD.  Fig.  3. 

As   the  plane  in  this  case  cuts  one  !«ide  i4  the  cone 
parallel  to  its  opposite  side,  it  is  evident  that  it  can  never 
meet  the  latter;  the  section  therefore  may  be  (  ontinued    i.  lo 
indefinitely,  if  the  cone  is  supposed  to  be  indefinitely  ex- 
pended. 

2 


■  CONIC  SECTIONS. 

G.  If  the  plane  cut  the  cone  obliquelj,  so  as  to  make  an 
&ngle  with  the  base  less  than  that  made  by  the  side  of  the 
^ 'e-  4.  cone,  the  section  is  called  an  Ellipse  :   as  ABC. 

As  the  cutting  plane  in  this  case,  is  not  parallel  with  the 
side  of  the  cone  V  K,  it  will  meet  that  side — that  is  it  will 
intersect  the  surface  of  the  cone  on  every  side.  This  sec- 
tion therefore  is  complete  and  definite  for  every  position 
of  the  plane. 

7.  If  the  cutting  plane  makes  with  the  base  of  the  cone 
Fie.  5.  ^"  angle  greater  than  that  made  by  the  side  of  the  cone, 
the  section  is  called  an  Hyperbola  ;  as  ACD, 

The  cutting  plane  in  this  position  will  never  meet  the 
opposite  side  of  the  cone  in  the  direction  of  V  K,  but  it 
will  meet  it  produced,  on  the  other  side  of  V,  and  if  all  the 
sides  of  the  cone  be  supposed  to  be  produced  through  the 
vertex  V,  forming  an  opposite  and  similar  cone,  and  if 
the  plane  also  be  produced  cutting  this  cone  in  the  section 
Bed,  the  two  sections  ACD  and  Bed  are  called  Opposite 
Hyperbolas, 

Tig.  6.  8.  If  there  be  four  cones,  the  angles  at  whose  vertices 
are  together  equal  to  four  right  angles  ;  and  if  their  axes 

].^  13.  be  all  placed  in  one  plane  and  their  vertices  all  meet  in  a 
given  point  V,  the  sides  of  the  cones  will  touch  each  oth- 
er in  the  right  lines  LVF,  BVH.  If  a  plane  per- 
pendicular to  that  in  which  the  axes  of  the  cone  are  pla- 
ced, and  parallel  to  O  P,  the  axis  of  the  two  cones  BVL 
FVH,cut  these  cones  in  the  opposite  hyperbolas  AD 
Bd  ;  if  another  plane,  also  perpendicular  to  that  in  which 
the  axes  of  the  cone  are  placed,  pass  through  A  or  B  par- 
allel to  KVM,  the  axes  of  the  remaining  cones,  cutting 
opposite  hyperbolas  in  these  cones,  these  two  pairs  of  op- 
posite hyperbolas  are  mutually  conjugate  to  each  other. 
The  two  sections  are  commonly  supposed  to  be  placed  in 
the  same  plane,  the  lines  AB  and  ab  bisecting  each  oth- 
er, as  in  fig.  7. 

If  the  angle  at  the  vertex  of  each  of  the  cones,  (fig,  6) 
be  a  right  angle,  the  cones  will  all  be  equal  and  similar, 
and  the  sections,  being  at  the  same  distance  from  the  ver- 
tex, will  be  equal  and  similar,  and  AB  will  be  equal  to  ab ; 
in  this  case  they  are  called  Equilateral  Hyperbolas,  and 
may  be  all  cut  by  one  plane,  parallel  to  that  in  which  th« 
axes  of  the  cones  are  placed  ;  as  represented  in  fig.  7. 


Cor. 


CONIC  SECTIONS.  7 

t).  As  the  properties  of  the  triangle  snd  circle  are  de- 
monstrated in  the  Elements  of  Euclid,  and  are  investiga- 
ted without  any  reference  to  the  sections  of  a  cone,  they 
are  not  usually  reckoned  among  the  conic  sections  ; — this 
term  being  appropriately  applied  to  the  three  remaining 
sections, — viz.  the  Parabola,  Ellipse  and  Hyperbola. 

10.  The  Vertices  of  a  conic  section,  are  the  points 
where  the  cutting  plane  meets  the  opposite  sides  of  the 
cone  ;  or  the  sides  of  a  vertical  triangular  section  through 
the  axis  of  the  cone,  and  perpendicular  to  the  plane  of  the 
given  section. 

In  the  Parabola,  the  plane  does  not  meet  the  opposite  ^^s-  3. 
side  of  the  cone  VH,  (3)  *  This  section  therefore  has  but 
one  vertex,  as  A  . 

The  Ellipse  has  two  vertices,  and  opposite  Hyperbolas 
have  two  ;  as  AB.  ^^'^'^'^' 

1 1.  The  Transverse  axis,  is  the  straight  line  which  con- 
nects the  vertices  ;  as  AB. 

This  definition  is  applicable  to  the  Ellipse  and  opposite 
Hyperbolas,  but  not  to  the  Parabola  which  has  but  one 
vertex.  The  axis  of  the  Parabola,  may  be  defined,  the 
intersection  of  its  plane  with  that  of  the  vertical  triangu- 
lar section  to  which  it  is  perpendicular.  This  definition  Sup. 2. 3. 
is  applicable  to  all  the  conic  sections,  since  it  evidently 
coincides  with  that  before  given  for  the  Ellipse  and  Hy- 
perbola. The  line  of  their  mutual  intersection  passes 
through  the  vertex  of  the  Parabola  and  is  of  indetermi- 
nate length  ;  since  the  section  itself  may  be  indefinitely 
continued. 

12.  The  Centre  of  a  conic  section  is  the  middle  of  the 
transverse  axis. 

Hence  the  Parabola  has  no  centre  ;  that  of  the  Ellipse 
is  within  the  section,  and  that  of  the  Hyperbola  without  it. 
between  the  opposite  Hyperbolas. 

13.  The  Conjugate  axis  of  a  conic  section,  is  a  line 
drawn  through  the  centre,  at  right  angles  to  the  transverse 
axis. 

*  This  figure  and  others  similar,  refer  to  the  articles  numbered  in  tljis 
^ieries  of  definifions. 


J  CONIC  SECTIONS. 

Hence  the  Parabola  has  no  conjugate  axis.  In  the  El- 
lipse it  is  within  the  section  and  terminated  by  the  curve. 
The  conjugate  axis  of  the  Hyperbola  is  without  the  sec- 
tion and  is  bounded  by  the  curves  of  the  conjugate  Hy- 
perbolas. Or  more  correctly,  the  conjugate  axis  is  the 
transverse  axis  of  the  conjugate  Hyperbolas.  (8)  Hence 
the  two  axes  are  mutually  conjugates  to  each  other. 

The  preceding  definition,  of  the  conjugate  axis  of  a 
Hyperbola,  is  correct,  in  all  cases  in  which  it  is  applicable. 
It  is  limited  however  in  its  application,  to  those  cases  in 
which  the  Hyperbola,  is  cut  by  a  plane  parallel  to  the  Ax- 
is of  the  cone. 

The  following  limitation  of  the  conjugate  axis  of  a  Hy- 
perbola is  general.  It  is  a  mean  proportional  between 
the  diameters  of  the  circular  sections,  which  pass  through 
the  vertices  of  the  transverse  axis  of  the  Hyperbola. 
Thus  the  conjugate  axis  of  A  B,  is  equal  to  a  mean  pro- 
*'»§;•  5.  portional  between  A  Q  and  B  R. 

14.  A  Diameter  of  any  conic  section,  is  a  straight  line, 
drawn  through  the  centre,  and  terminated  in  both  direc- 
tions by  the  curve. 

This  definition  of  a  diameter  evidently  includes  the 
axis. 

This  definition  is  inapplicable  to  the  Parabola  which 
has  no  centre.  (12).  A  diameter  of  a  Parabola,  may  be 
defined,  any  line  drawn  parallel  to  its  axis^  and  terminated, 
like  the  axis,  in  one  direction  by  the  curve,  while  the  oth^ 
er  may  be  indefinitely  extended,  as  Cd. 

15.  The  extremities  of  a  diameter,  or  its  intersections 
with  the  curve  are  called  its  Vertices, 

16.  The  Conjugate  to  any  diameter,  is  a  line  drawn 
through  the  centre  of  the  section,  and  parallel  to  the 
tangent  *  at  the  vertex  of  the  diameter.     7^hus  D  E  is  the 

Pi^  8.  conjugate  of  the  diameter  C  H. 

This  definition  evidently  includes  the  conjugate  axis 
(13.)  If  it  is  admitted,  or  assumed,  that  the  tangent  at  the 
vertex  of  the  transverse  axis,  is  perpendicular  to  it. 

*  Euclid's  definition  of  a  line  touching  a  circle,  may  be  applied  to  a  tan- 
gent of  a  Conic  Section.  It  is  a  line  which  meets  the  curve,  but  does 
not  cut  it. 


Fi-.  9. 


CONIC  SECTIONS. 


17.  An  Ordinate  to  any  diameter  is  a  right  line,  parallel 
to  the  tangent  at  its  vertex,  and  terminated  in  one  direc- 
tion by  the  curve,  and  in  the  other,  by  the  diameter  as 
K  G.     Figures  8  and  9. 

1 8.  The  Jlbsciss  to  any  ordinate  is  the  part  of  the  diam- 
eter, contained  between  its  vertex  and  that  ordinate,  as  ^^S-  ^^ 
C  G  and  G  H.  "^  ^• 

Hence  in  the  Elhpse  and  Hyperbola,  every  ordinate 
has  two  determinate  abscisses.  In  the  Parabola,  there  is 
but  one.  the  other  being  supposed  indefinitely  extended. 
(ll.)asGJ.      Fig.  9. 

19.  The  Parameter  of  any  diameter,  sometimes  called 
the  Lotus  Rectum,  is  a  third  proportional  to  that  diameter 
and  its  conjugate. 

This  detinition  is  not  applicable  to  the  Parabola,  (13.) 
but  its  Param(iier  is  a  third  proportional  to  any  absciss 
and  its  ordinate. 

20.  The  Focus  of  a  conic  section,  is  the  point  in  the 
axis  where  the  ordinate   is  equal  to  half  the   parameter. 

As    F  whore  C  F  is  equal   to  the  semi-parameter  of  the  fj?-  7. 
section.  !i.&9. 

The  Ellipse  and  Hyperbola  have  each  two  Foci  as  F, 
/,  the  Parbola  has  but  one. 

21.  If  tangents  be  drawn  to  the  four  vertices  of  the 
curves,  in  Conjugate  Hyperbolas,  forming  an  inscribed 
rectangle,  thediagonals  of  this  rectangle,  are  called  the 
Assymptotes  of  the  Hyperbolas.    As  L  F  and  K  H.  p.^  -, 

22.  Similar  conic  sections,  are  those  of  the  same  kind, 
whose  transverse  and  conjugate  axes  have  to  each  other 
the  same  ratio. 

As  this  definition  is  not  applicable  to  the  Parabola,  Sijn- 
ilar  Parabolas^  are  those  whose  abscisses  and  ordinates  are 
to  each  other  in  the  same  ratio. 

23.  When  a  magnitude,  which  is  supposed  to  vary  ac- 
cording to  a  certain  law,  approaches  continually  towards 
equality  with  another  given  magnitude,  so  that  ultimately 
the  former  differs  from  the  latter  less  than  by  any  assign- 


i^  CONIC  SECTIONS 

able  magnitude,  then  the  latter  is  said  to  limit,  or  to  ht 
the  limit  of  (he  former,  and  their  ultimate  or  limiting  ra- 
tio, is  said  to  be  a  ratio  of  eqnahty. 

Thus  Euclid,  (Book  XI 1  Prop.  2.)  dennonstrates  that  i 
a  regular  polygon  iriscribt;d  in  a  circle,  will,  as  the  num- 
ber of  its  sides  is  supposed  to  be  increased,  approach  con- 
tinuall)  to  equality  with  the  circle,  so  as  ultimately  to  dif- 
fer from  the  circle  less  than  by  any  assignable  magnitude.  I 
By  this  truth,  he  demonstrates,  that  the  areas  of  circles 
are  as  the  squares  of  their  diameters. 

In  this  example,  the  circle  is  called  the  limit  of  the  pol- 
ygons. He  also  demonstrates  that  in  this  sr*.nse,  the  circle 
is  also  the  limit  of  polygons  circumscribed  about  it. 

*'  The  consideration  of  limits  more  or  less  dis^juised, 
must  unavoidably  enter  into  every  investigation  wh»ch  has 
for  its  object,  the  mensuration  of  the  circle."       Leslie, 

"This  principle  is  general,  and  is  the  only  one  by  which 
we  can  possibly  compare  curvilineal  with  rectilineal  spa- 
ces, or  the  length  of  curve  lines  with  the  length  of  straight 
lines,  whether  we  follow  the  methods  of  the  ancient  or  of 
the  modern  geometers."  Playfair. 

General  Scholium, 

The  cutting  plane  being  parallel  to  the  base,  forming  a 
circular  section,  if  it  be  supposed  to  revolve  about  A,  at 

Fig.  10.  the  commencement  of  its  motion,  when  it  deviates  in  the 
least  pos'^ible  degree  from  parallelism  with  the  base  of  the 
cone,  the  section  will  be  an  Ellipse,  which  in  the  language 
of  mathematicians  is  infinitely  near  to  a  coincidence  with 
the  circular  section.  That  is,  it  may  differ  from  the  circle 
less  than  by  any  assignable  difference. 

If  now  the  plane  b*-  supposed    to   revolve  from  B'   to- 

t  J-.  10.  ^aj.(jg  B''-in  passinjr  through  B,  where  it  is  parallel  to 
the  base,  the  circle  will  be  the  section  which  forms  the 
transition  from  the  Ellipse  immediately  on  <ne  side  of  it, 
to  that  immediately  on  the  other.  It  is  the  limit  towards 
which  each  Ellipse  approaches  indefinitely,  and  may 
therefore  be  considered  an  Ellipse,  whose  transverse  and 
conjugate  axes  are  equal,  ana  whose  Foci  unite  in  the 
centre. 

Again — if  the  plane  be  supposed  to  revolve  towards  a 
parallelism  with  the  opposite  side  of  the  cone,  the  trans- 
Terse  axis  oi  the  Ellipse  becomes  extended,  and  the  sec- 


CONIC  SECTIONS.  H 

tion  approaches  towards  that  of  a  Parabola,  from  which, 
iminediately  before  the  plane  becomes  parallel  with  the 
side  of  the  cone,  it  differs  less  than  by  any  assignable  dif- 
ference. In  other  words,  the  Parabola  is  the  limit  to 
which  the  Ellipse,  by  the  revolution  of  the  plane,  contin- 
ually and  indefinitely  approaches. — The  Parabola,  there- 
fore, has  been  called  an  Ellipse  whose  axis  is  infinite,  and 
whose  centre  is  at  an  infinite  distance. 

Again — immediately  after  the  plane  ceases  to  be  paral- 
lel to  the  side  of  the  cone,  the  section  is  an  Hyperbola, 
which  is  indefinitely  near,  in  resemblance  and  coincidence, 
to  the  Parabola,  which  is  the  limit  to  which  the  Hyperbola 
approaches  ;  it  may  therefore  be  called  a  Hyperbola 
whose  axis  is  infinite.  The  Parabola  is  therefore  the 
limit  between  the  Ellipse  and  the  Hyperbola,  and  holds 
the  place  of  transition  from  one  to  the  other. 

Lastly — If  a  Hyperbolic  section,  perpendicular  to  the 
base  of  the  cone,  be  supposed  to  move  parallel  to  itself,  it 
will  at  one  point  coincide  with  the  axis  of  the  cone.  The 
section  will  then  become  a  triangle,  (def.  2.)  Immedi- 
ately before  and  after  this  coincidence,  the  section  will  be 
a  Hyperbola,  which  is  indefinitely  near  to  the  triangular 
section.  The  latter  is  the  limit  to  which  the  Hyperbolic 
sections  approach  indefinitely  on  each  side — it  is  the 
transition  from  one  to  the  other,  and  may  therefore  be 
considered  as  a  H}  perbola  whose  axis  is  infinitely  small,  or 
evanescent,  and  whose  Foci  unite  with  their  vertices  in  the 
vertex  of  the  triangle.  The  sides  of  this  triangular  sec- 
tion (which  is  the  limit  of  the  Hyperbolas,  on  either  side 
of  it,)  evidently  coincides  with  the  tvio  assy mptotes  of  the 
Hyperbolas. 

From  the  connexion  and  intimate  relations  thus  seen  to 
exist  between  all  the  Conic  Sections,  we  should  ex- 
pect striking  resemblances  and  analogies  in  their  proper- 
ties. It  will  be  one  object  in  the  following  treatise  to 
point  these  out  as  they  occur. 

The  most  simple  figures  produced  by  the  intersection  of 
a  cone  and  plane,  are  the  triangle  and  circle  ;  but  for  the 
reasons  before  mentioned  (9)  they  are  here  omitted,  and 
the  properties  of  the  Parabola,  the  section  next  in  sim- 
plicity to  them,  will  be  first  examined. 


OF  THE  PARABOLA. 


PROPOSITION  I. 

The  squares  of  any  two  ordinates,  are  to  each  other  as 
their  abscisses. 

Fig.  11.  Let  NVM  be  a  triangular  section  through  the  axis  of 
the  cone,  AGIH,  a  ParaboHc  section  made  by  a  plane, 
perpendicular  to  the  fornier,  AH  the  line  of  their  niutual 
intersection,  will  be  the  axis  of  the  Parabola,  (1 1,)  also 
let  there  be  two  circular  sections  parallel  to  the  base  of 
the  cone,  and  cutting  AH  in  the  points  F  and  H;  FG  and 
HI,  the  intersections  of  these  planes  with  that  of  the  Para- 
"P*  •  'bolic  section,  will  be  perpendicular  to  the  plane  of  the  tri- 
angular section,  and  therefore  perpendicular  to  HA,  and 

Sup.  2.  toFL  and  HN ;  they  are  therefore  ordinates  (17)  both  in 
^'  the  Parabola  and  in  the  circles. 

The  proposition  then  is    FG-  :  HP  : ;  AF  :  AH 

6.  4.    By  sim.  tri.  AF   :  AH  : :  FL  :  HN; 

but,  because  AH  is  parallel  to  VM  and  KL  to  MN, 

1-  34.  .-.  KH  is  a  parallelogram  and  KF=MH, 

6^     therefore  AF  :  AH:  :KF .  FL  :  MH.  HN  ; 

V^Sf/but  inthe  circle  FG2=KF.FLand  HP=MH.HN, 

5  7    therefore  FG^  :  HP  : :  AF  :  AH. 

Q.  E.  D, 

Cor.  1. — This  is  true  of  ordinates  on  both  sides  of 
the  axis,  as  the  demonstration  is  equally  applicable  lo  or- 
dinates on  either  side;  and  the  ordinates  in  the  circle  are 
evidently  bisected;  therefore  FS=FG,  or  GS  is  double 
of  FG,  it  is  therefore  called  a  doyhle  ordinate  ;  also  the 
curve  on  each  s'de  of  the  axis  is  similar,  the  whole  Parab- 
ola is  bisected  by  the  axis,  and  the  two  parts,  if  laid  one 


OF  THE  PARABOLA.  13 

upon  the  other,  will  coincide  in  every  respect — for  il  they 
did  not  coincide,  in  any  part,  the  corresponding  ordinates 
on  opposite  sides  of  the  axis,  would  be  vneqiiaL  For  a 
similar  reason,  the  two  parts  of  the  tangent  at  the  vertex, 
will  coincide; — they  therefore  make  equal  angles  with 
the  axis — in  other  words,  the  tangent  at  the  vertex  of  a 
Parabola,  is  perpendicular  to  the  axis. 

ScHOL. — If  any  straight  line  FG,  be  supposed  to  move 
parallel  to  itself,  upon  another  strai-ht  line,  as  AH,  to 
which  it  is  perpendicular,  and  if  the  line  FG  vary  in  its 
length  continually,  so  that  its  squares  have  always  the  same 
ratio  to  each  other  as  its  distances  from  the  point  A,  the 
'line  FG  will  describe  the  half  of  a  parabolic  section.  Or, 
(as  FG2  equals  the  rectangle  GF.FS)  if  any  line,  as  GS, 
be  bisected  by  AH,  to  which  it  is  perpendicular,  and 
moving  parallel  to  itself,  vary  in  such  a  manner  that  the 
rectangles  of  its  parts  are  as  its  distances  from  A,  it  will  de- 
scribe a  Parabola.  And,  because,  when  the  absciss  in- 
creases, the  rectangle  GF  .  FS  increases  in  the  same  ratio, 
the  curve  will  always  recede  from  the  axis;  but  as  this 
rectangle,  or  the  square  of  FG,  increases  more  rapidly 
than  FG  itself,  or  in  the  duplicate  ratio  of  FG,  the  curve 
will  recede  less  for  any  given  increment  of  the  absciss, 
as  it  is  more  distant  from  the  vertex,  or,  in  other  words, 
it  is  always  concave  to  the  axis. 

Cor.  2.  (19)       AF  :  FG:  :FG  :  P=:Parameter, 
therefore  FG3=AF.P,  6.17. 

also  (19)  AH  :HI::Hl  :  P', 

and  HI-^=AH.P',  Con- 

therefore  FG-  :  HI^ ; ;  AF  .  P  :  AH  .  P',  ^7^^/^ 

but  (1)  FG^  :  HI-^::AF:  AH,  6  i 

therefore  AF  :  AH:  :AF  .  P  :  AH.  P.  5.  n. 

P=P',  or  the  Parameter  of  the  axis 
is  a  constant  quantity  ;  also  the  rectangle  of  this  constant 
quantity,  and  any  absciss  of  the  axis,  is  equal  lo  the  rec- 
tangle of  the  equal  jsarts  of  the  double  ordinate, — that  is, 
to  the  square  of  the  ordinate.  This,  expressed  in  alge- 
braical language,  is  called  the  equation  of  the  curve,^ 

*  Day's  Algebra— Sec.  22— Let  the  Parameter  be  represented  by  a,  tho 
absciss  by  I,  and  the  ordinate  by  ,v ;  then  ax-yZ:  this  is  termed  "  tljG 
equation  olthe  Curve,"  of  a  Parabola. 

o 

O 


14  OF  THE  PARABOLA. 


ScHOL — From  this  equality  between  the  square  of  any 
ordinate,  and  the  rectangle  of  its  absciss  and  parameter, 
Appollonius  of  Perga.  called  this  section  the  Parabola, 
{Appollonii  Conica,  per  J.  Barrow  10  page.  Lon,  1675,  alsn 
Hamiltoii^s  Conic  Sections  84.) 


PROPOSITION^  11. 

If  any  double  ordinate  to  the  axis,  intersect  any  num- 
ber of  diameters,  the  parts  of  the  diameters  cut  off  by  the 
double  ordinate,  are  to  each  other,  as  the  rectangles  of  the 
segments  into  which  they  divide  the  double  ordinate. 
*'•?•  12.  That  is,      AH  :  GL:  :KH  .  Hi  :  KL  .  LI. 

For(I.  Cor.  2.)P  .  AH  =  HI2, 
3.ax&  and  P  .  AF=FG2, 

1,6.         tlierefore       P  .  (FH=)  GL=Hl3 -FGS 
t.6  Cor.  or  P  .  GL=KL  .  LI, 

5.  14.  therefore   P  .  AH  :  P  .  GL:  :(Hl2=)KH .  IH  :  KL.LI; 

6.  1.  or  AH  :  GL:  :KH  .  HI  :  KL  .  LL 

Q.  E.  D. 

The  same  demonstration  i^  evidently  applicable  to  any 
other  diameter,  as  ED. 

ScHOL. — If  then  any  line,  as  AH,  be  supposed  to  move 
parallel  to  itself,  on  KI,  to  which  it  is  always  perpendicu- 
lar, and  to  vary  in  its  length  continually,  in  the  same  ratio 
as  the  rectangles  into  which  it  divides  KI,  the  space  descri- 
bed by  it  will  be  a  parabola;  whose  highest  point  or  ver- 
tex will  be  above  the  middle  of  the  base,  because  the  rec- 
6.  27.  tangle  is  the  greatest  when  the  line  is  equally  divided. 

Cor.  1. — As  this  is  equally  applicable  to  all  double  or- 
dinates  to  the  axis,  the  proposition  may  be  stated  univer- 
sally, the  parts  of  all  diameters  cut  ojfhy  any  double  ordinate 
to  the  axis^  are  to  the  parts  cut  off  by  any  other  double  ordin- 
ates,  as  the  rectangles  of  the  segments  into  which  they  divide 
the  former  double  ordinate,  are  to  the  rectangles  of  the  seg- 
ments into  zohich  they  divide  the  latter* 

That  is    GL  :G/::KL  .  LI  iH  .  li. 
and       GL  :  EdwKL  .  LI  :  kd ,  di. 


OF  THE  PARABOLA 


15 


This  proposition  evidently  includes  Proposition  1.  Also 
in  all  cases,  the  rectangle  of  the  segments  of  any  double 
ordinate  to  the  axis,  is  equal  to  the  rectangle  of  tl  e  Para- 
meter of  the  axis,  and  the  part  of  the  diameter  which  is 
cut  off  by  the  double  ordinate. 

That  is  P  .  GL=KL  .  LI;  and  P  .  ED=KD  .  DI.  (See 
dem.  of  Prop.  2d.) 

This  also  includes  Cor.  2,  Prop.  I.  6.  1«. 

Cor.  2.     As  the  Parameter  of  the  axis, 

To  the  sum  of  any  two  ordinates. 

So  is  the  difference  of  those  ordinates, 

To  the  difference  of  their  abscisses. 

For  the  last  step  but  two  of  the  demonstration  is, 

p:gl=kl  .Li. 

therefore  P  :    KL(orHl  -f  FG):  :LI(or  HI-FG)  :  GL. 


PROPOSITION  III. 

If  any  double  ordinate  be  produced  and  intersect  any  num- 
ber of  diameters  icithout  the  curve,  the  external  segments 
of  these  diameters,  are  to  each  other  a  he  rectangles  of 
the  corresponding  secmentsof  the  line. 

That  is  nl  :  6G  :  gn  .  no  :  gb  .  bo.  Fi-  12. 

For  (1)  (Fm2=)  Hl»  =  AH  .  P, 
and      (fb^  =:)  FG=  =  AF  .  r 

HI2-     FG2(  =  Gm.mN)=FH.P=ml  .P,      2-5. 
and      FG2  -     Fg^  (=  ^6  .  bo)=       ¥f  ,  P=6G  .  P  ; 
therefore  Gm  .  mN  :  gb  .bo'.'.ml  .  V  \  bG  .  ?:\m\  :  ^G  ; 
so        gn  .no     \  gb  ,  bo\'.n\  \  bG 
or         72I    :  6GI  '.gn  ,  no  :  gb  ,  bo. 

Q.  E,  D. 

Cor.  I.     Hence  gb  .  bo  :  Gni  .  tmN:  i^G  :  m). 

CoR.  2.  \Yhenfo,  approacl^es  Aa,  th  rectangle  gn  . 
no,  approaches  continually  and  indefi'ittely  t  wards  equal- 
ity with  the  square  o  Aa,  the  :::icrcepted  pait  of  the 
tangent  at  the  vertex;  when  thcrftfore/  coir-rides  with  A, 
the  property  becomes  Gm  .  mN  :  Aa- :  '.ml  :  al 
and  gb  .  0  :  Ah^::bG  :  kG 
An^  :  A;?2  ..^i  :  ;^G 


j.g  OF  THE  PARABOLA. 

That  is,  the  squares  of  the  intercepted  parts  of  the  tan-- 
gent  at  the  vertex,  are  as  the  external  parts  of  the  diame- 
ters intercepted. 


PROPOSITION  IV. 

If  a  tangent  be  drawn  to  any  point  of  the  Parabola 
meeting  the  axis  produced  ;  and  if  an  ordinate  to  the  ax- 
is be  drawn  from  the  point  of  contact  ;  then  the  absciss 
of  that  ordinate  will  be  equal  to  the  external  part  of  the 
axis. 

That  is,  if  T  C  touch  the  curve  at  the  point  C,  and  CM 
Fig.  26.  be  an  ordinate  to  the  point  of  contact, 
then  AT=:AM 

For  from  the  point  T  draw  a  line  cutting  the  curve  in 
the  two  points  E,  H;  to  which  draw  the  ordinates  D  E, 
GH; 


Then  by  (I.)  AD 
tt   and  by  sim.  tri.  TD^ 
5^  ft.        therefore       AD 
5.  A.  17.           and            AD 

:  AG  ::DE2 
:TG^::DE2 
:  AG  ::TD2 
:DG  ::TD2 

:GH% 
:GHs 
:TG% 
:  TG2-TD% 

2.5.  Cor.             but             TG2- 

-TD2=:DG 

.  TD  +TG, 

5.  7.               .-.            AD 

6.  1.    and  the  reel's.  AD 

:  DG  ::TD2 
.  TD    : TD 

:  DG    .  TD+TG, 
.  DG::TD2  :  DG 

5   ig       consequently    AD 

g   1     therefore  the  base?  AD 
S.A.&E.           and            AD 

6.  16.                  „                AD 
5.A.&E.                „                AD 

TD+TG, 
.  TD    :  TD2 

TD+TG, 
:  TD  ::TD 
:  AT  ::TD 
:TD  ::AT 
:  AT  ::AT 

::TD.DG    :  DG 

:  TD  +TG, 
:TG, 
:TG, 
:  AG. 

or  AT  is  a  mean  proportional  between  AD  and  AG, 

Now  if  the  line  T  H  be  supposed  to  revolve  about  the 
point  T,  then  as  it  recedes  from  the  axis  and  approaches 
the  tangent  T  C,  the  points  E  and  H  approach  the  point 
of  contact,  and  when  T  H  coincides  with  the  tangent, 
the  points  E  and  H  will  unite  in  C.  But  when  the  points 
E  and  H  approach  towards  each  other  in  C;  D  and  G  also 
approach  towards  each  other,  and  towards  M.  Conse- 
quently A  D  and  A  G  approach  towards  the  ratio  of  equal- 


IG. 


OF  THE  PARABOLA.  17 

ity  ;  and  before  they  unite  in  M,  they  differ  from  each 
other  less  than  by  any  assignable  quantity  ;  their  luniting 
ratio,  therefore,  in  M  is  a  ratio  of  equahty.  But  AT  is 
always  a  mean  proportional  between  them,  it  is  always 
therefore  greater  than  AD  and  less  than  A  G,  and  like 
A  M  is  a  limit  between  A  D  and  A  G  ;  it  is  therefore 
equal  to  AM.  That  is,  the  external  part  of  the  axis  cut 
off  by  the  tangent  is  equal  to  the  absciss  of  the  ordinate 
to  the  point  of  contact. 

Q.E.D. 

Cor.  1 . — AM  or  AT  is  a  mean  proportional  between 
AD  and  AG. 

Cor.  2.  As  this  is  true  of  a  tangent  drawn  from  either  Fi^ 
extremity  of  a  double  ordinate,  the  tangents,  CT,  DT 
intersect  the  axis  in  the  same  point,  and  make  equal  an- 
gles with  it.  And  conversely,  a  line  bisecting  the  angle 
made  by  the  tangents,  or  a  line  drawn  from  that  point  and 
bisecting  the  double  ordinate,  is  the  axis. 

Cor.  3. — If  two  or  more  Parabolas  have  the  same  axis 
and  vertex,  and  any  ordinate  in  one  section  be  produced 
until  it  cut  the  other,  and  tangents  to  the  curve  be  drawn 
from  the  points  of  intersection,  these  tangents  will  all 
meet  the  axis  in  the  same  point. 

Since  in  each  AT  =  AM. 

ScHOL. — From  this  proposition  it  is  easy  to  draw  a  tangent 
(o  the  curve  from  any  point,  either  in  the  ourve  or  in  the 
axis  produced.  If  it  be  in  the  curve,  as  C,  draw  an  ordi- 
nate from  the  point,  and  take  a  point  T  in  the  axis  produ- 
ced, at  a  distance  from  the  vertex  equal  to  the  ordinate;  a 
straight  line  connecting  this  point  with  the  given  point, 
will  be  the  tangent. 

If  the  given  point  be  in  the  axis  produced,  take  an  ab- 
sciss AD,  equal  to  the  distance  of  the  point  from  the  ver- 
tex, the  ordinate  to  this  absciss  will  cut  the  curve  in  the 
point  of  contact ;  the  line  which  connects  this  with  the 
given  point,  will  be  the  tangent. 

CoR.  4. — The  subtangent,  TM,  is  equal  to  twice  the  ab- 
sciss. 


II  OF  THE  PARABOLA. 


PROPOSITION  V. 

If  there  be  any  tangent  and  double  ordinate  to  the  axis, 
drawn  from  the  point  of  contact,  and  also  any  other  diam- 
eter limited  by  the  tangent  and  double  ordinate,  tfien  shall 
the  curve  divide  that  diameter  in  the  same  ratio  as  the 
diameter  divides  the  double  ordinate. 
Fig.  13..      That  is       I E  :  EK : :  CK  :  KL. 


6.  4.    Forbvsim.tri.CK  :  KI  : 

:CD  :  DT  or2DA, 

but(def.  19.)     P:CD: 

:CD  :  DA, 

5. 4.  Cor.         and             P  :  CL: 

:CD  :  2DA, 

6.  22.       therefore        P  :  CK: 

:CL:  KI; 

but  (11  Cor.  2.)   P  :CK: 

:KL: KE; 

5.  11.       therefore     CL  :  KL: 

:KI   :  KE; 

5.17.           and         CK:KL: 

:IE   :  EK; 

or          IE  :  EK. 

:CK  :  KL. 

Q.E.D. 


14.      Cor.  1.— If  CK=KL,  then  IE=EK. 


PROPOSITION  VL 

The  same  being  supposed  as  in  the  last  proposition,  the 
external  parts  of  the  diameters,  between  the  curve  and 
tangent,  are  to  each  other  as  the  squares  of  the  intercepted 
parts  of  the  tangent. 

Fig.  13.  That  is,  IE  :  TA:  :CP  :  CT^, 

For  (V)  IE  :EK::CK:KL, 

^'  1   And  IE  :EK::CK2  :  CK  .  KL; 

In  the  same  way  TA  :  AD:  iCD^  :  CD  .  DL; 

But  (\l)  EK  :  AD:  :CK  .  KL  :  CD  .  DL, 

.  oo    Therefore  IE  :  TA:  :CK^  :  CD^; 

VI  And  IE:TA::CI^  :CT2. 

Q.  E.  D, 


OF  THE  PARABOLA. 


PROPOSITION  VIL 


19 


The  distance  from  the  Focus  to  the  vertex,  is  equal  to 
one  fourth  of  the  Parameter  of  the  axis. 
«r  AF=iP  Fig.  14. 

for  (19)      AF:FL::FL:P 

but  (20)      FL=iP.  .AF=iFL=iP  5.  c 

Q.  E.  D. 


CoR.  1. — If  a  tangent  be  drawn  to  the  curve,  from  the 
point  L,  it  is  called  the  Focal-tangent,  and  the  distance  of 
its  intersection  with  the  axis  from  the  Focus,  equals  the  fo- 
cal-ordinate.     That  is  F/=(2FA=)  FL. 

Cor.  2.  —  Also  the  distance  of  any  point  in  this  tangent 
from  the  Axis,  equals  the  sum  of  the  focal  distance,  and 
the  corresponding  absciss,  that  is  0N  =  N<=NA4-A^= 
NA-f  AF. 

Cor.  3. — The  tangent  at   the  vertex,  AP,  intercepted    6«  <i 
by  the  Focal-tangent,  is  equal  to  half  FL,  (theordinate  to 
the  Focus, )  =  iP  =  AF=A/,   so  that  A  is  the  centre  of  a 
circle  passing  through  t,  P  and  F. 

CoR.  4.  — If  through  the  point  T,  where  the  Focal-  ^^S'  ^ 
tangent  intersects  the  axis,  a  perpendicular  to  the  axis  be 
drawn  it  is  called  the  Directrix,  and  it  is  evident  that  the 
distance  of  any  point  of  the  curve  as  M  from  the  Direct- 
rix is  equal  to  ON,  or  the  ordinate  drawn  through  this 
point,  and  produced  to  meet  the  Focal-tangent. 

For  MB=OT=ON=(Cor.  2.)  N«=(Fig.  14.)NA+ 
AF. 


2b  OF  THE  PARABOLA. 


PROPOSITION  VIII. 


If  a  tangent  to  the  curve  nneet  the  axis  produced,  thctt 
the  distance  of  the  Focus  from  the    point  of  contact,    is 
equal  to  the  distance  of  the  Focus  from  the  intersection  of 
the  tangent  and  axis. 
Fig.  14.     That  is         FC=iFT. 

For  draw  the  ordinate  CD,  to  the  point  of  contact, 
then  (IV.)    AT=AD, 
Ax.  2.     therefore       FT=AF4-AD, 
2-  4.  and  FT2  =  AF^  +  AD^  +2AF.  AD; 

147  again         FC2=FD2 -f  CD^, 

but  FD=AD-AF.  and  FD2=AF3+AD^ 

*-  ^AF.  AD; 
and  (I  Cor.  2.)Cb2  =P.  AD=4AF.  AD     (VII) 
FC2=AF2  +  AD2-i-2AF.AD. 
'^^^  ^'      therefore     FT2=FC2  and  FT=FC. 

Q.  £.  D. 


Cor.   1. — The  triangle  FCT,  is  isosceles,  and  the  angle 
'•  ^'    formed  by  the  tangent,  and   a  line  drawn  to  the  Focus,  is 
equal  to  that  formed  by  the  tangent  and    the  the  axis^  or 
1.  29.    any  diameter  ;  that  is  FCT=FTC=KCiM. 

Cor.  2. — If  CG  be  drawn  perpendicular  to  the  tangent 
at  C,  it  is  called  the  normal^  and  FG,  the  distance  of  its 
intersection  with  the  axis  from  the  Focus  equals  FC,  which 
by  the  proposition  equals  FT. 

For  draw  FH,  perpendicular  to  TC,  it  will  bisect  TC, ; 
therefore  HF  also,  bisects  TG,  consequently  FG  =  FT  = 
FC. 


1.  11. 

6.  2. 


9. 


CoR.  3. — Therefore  F  is  the  centre  of  a  circle  passing 
through  T,  C,  G,  and  the  angle  at  the  centre  TFC  is 
double  TGC ;  censequently  FCK  is  double  GCK  ;  or 
FCG=GCK. 

ScHOL. — It  is  a  law  in  optics,  that  the  angle  made  by  a 
ray  of  reflected  light,  with  a  perpendicular  to  the    reflecr 


OF  THE  PARABOLA. 

ting  surface,  is  equal  to  the  angle  which  the  incident  ray 
makes  with  the  same  perpendicular.  Hence,  if  a  ray  should 
fall  upon  the  concave  surface  in  the  direction  of  KC,  it 
would  be  reflected  in  CF;  and  all  parallel  rays,  since  they 
would  coincide  with  diameters,  would  be  reflected  to- 
wards the  same  point,  which  is  therefore  called  the  Focus. 

Cor.  4. —  The  subnormal  DG  equals  the  semi-param- 
eter, or  the  ordinate  at  the  Focus. 

For  (IV)    (TD=)2AD  :  DC.  :DC  :  DG,  6.  8 

but  (def  19)  AD  :  DC:  :DC  :  P  =  parameter. 
therefore      2AD  :  AD:  :P  :  DG;  ^-  ^• 

DG  =  iP.  5.  c 

also  GX  being  drawn  perpendicular  to  CF, 

CX  =  DG  =  iP;   for   the  triangles  CDG, 
GXC  are  similar  and  equal. 

Cor.  5. — A  perpendicular  to  the  axis  drawn  from  the 
vertex,  meets  the  tangent  in  the  same  point,  H,  in  which  a 
perpendicular  from  the  Focus  upon  the  tangent  meets  it. 
(See  Cor.  2.) 

Foras(IV)  AT=AD    .TH=HC.  6.2. 

Hence  also  AH  is  a    mean    proportional  between   AF 
and  AT,  that  is  between  AF  and  AD,the  distances  of  the  Fo- 
cus and  the  ordinate  to  the  point  of  contact,  from  the  ver-    6.  8 
tex; 

Or  AF  :  AH  :  AT.  are  continued  proportionals. 

Also,  FH,  is  a  mean  proportional  between  FA  and  FT, 
the  distances  of  the  Focus,  from  the  vertex  and  the  point 
©f  intersection  of  the  tangent  and  axis  produced. 

Or  FA  :  FH  :  FT,  are  continued  proportionals. 

Cor.  6.— -The  two  tangents  CHT  and  AHR  mutually 
bisect  each  other.  For  as  TA=iTD.'.AH  =  iDC  = 
lAR. 

Cor.  7. — Hence  FI,  a  line  drawn  from  the  focus  per- 
pendicular to  the  tangent  and  produced  to  meet  the  diam- 
eter which  passes  through  the  point  of  contact,  is  bisected 
by  the  tangent; — and,  conversely,  aline  drawn  from  C  bi- 
secting FI,  at  right  angles,  is  a  tangent  to  the  curve  at  that 
point. 

4 


2f 


22  OF  THE  PARABOLA. 

Cor.  8— Hence  (2nd  step,  dem.)  FT  (=FC)=FAH- 
AT=FA  +  AD.  \ 

f 

ScHOL.  From  this  property  the  curve  is  easily  descri- 
bed by  points. 

Let  A  be  the  vertex  of  the  axis,  draw  any  number  of 
lines  as  EE  perpendicular  to  the  axis,  then  with  the  dis- 
tances, TD  TF,  TO  as  radii,  and  the  Focus  as  a  centre, 
describe  curves  crossing  thr  parallel  lines  in  EE,  and  then 
draw  the  curve  through  all  the  points  E,  E,  E,  &:c. 


PROPOSITIOX  IX. 

If  tangents  be  drawn  to  the  vertex  of  the  axis,  and  of  any 
other  diameter,  meeting  the    axis  and  diameter   produced, 
the  triangles  thus  formed  are  equal. 
Fig.  20.        That  is,       AHT=CHE. 

For,  since    IV)TA=AD=EC,  and  AH=HE; 

Therefore  the  triangles  are  similar  and  equal. 

Cor.  l._If   AM  be  parallel  to  CT,  the   triangle  CDT 
1,36  &  4i.=AEM=CD.  DA=MDAC=MATC. 

CoR.  "2. — If  PI^^  be   perpendicular  to   AR,  and    PRL 
parrallel  to  CT.     Then  the  triangle  PIR  =  IA  .  AE. 
For  the  triangles  CDT    PIH   are  similar; 
6.  19.        Therefore— triangle  CDT  :  PIR:  iCb^  :  PI-% 
6  1  (I)  ::DA:1A, 

1  tr4i  ::DA  AE  :  lA  .  AE; 

But  CDT=DA  .^E.  .PIR-^IAAE. 

CoR.  3.— Hence  to  both  PIR  and  lA  .  AE,  add  LRIN, 
and  LPN  =  LRAE=LRTC. 

CoR.  4. — In  a  similar  manner  it  may  be  shown, 
thai,|?  i  R=iA  .  AE=TC  n  i. 

From  /?  1  R,  and  TC  n  i.  take  RL  n  i, 
andj9  n  L=LRTC  =  LPN. 

CoR.  5. — Tlierefore  P/?,  parallel  to  CT.  is  bisected  in 
L;or  every  diameter  bisects  its  double  ordinate. 


OF  THE  PARABOLA.  03 


SCHOLIUM, 


The  preceding  proposition  and  the  fifth,  enable  us  to 
pass  from  the  axis,  to  the  other  diameters  of  the  section, 
and  to  prove  the  same  things,  concerning  each  of  them, 
which  have  been  demonstrated  concerning  the  axis,  I'he 
preceeding  propositions  therefore,  except  that  which  re- 
fers to  the  Focus,  will  appear  to  he  particular  truths,  which 
are  all  included  in  the  more  general  propositions  that  fol- 
low. 

Care  has  been  taken,  that  these  ;o-enera/  properties  of  all 
the  diameters  of  the  Parabola,  be  arranged  in  the  same  or- 
der, and  expressed  in  the  same  form,  in  which  the  particu- 
lar properties  of  the  axis  have  been  ;  not  only  to  assist  the 
memory  of  the  student,  but  to  present  more  clearly  the 
universal  and  striking  analogies  which  are  observabale  in  the 
different  parts  of  the  Parabola. 


PROPOSITION  X. 

The  squares  of  the  ordinates  of  any  diameter,  are  as  its 
abscisses. 

That  is  QE^  :  RA^ :  :CQ  :  CR.  Fi-.  i 

This  lias  already  been  proved  concerning  the  axis,  (Prop. 
1  ;)  to  shew  that  it  is  al  so  true  of  any  other  diameter,  as 
CS,  draw  the  tangent  CP.and  the  externals  El,  AT,  paral- 
lel to  the  axis,  or  to  the  diameter  CS; 

Then  because  the  ordinates  QE,  RA  are  parallel  to  the 
tangent  CP,  by  the  definition  of  them,  therefore  the  figures 
IQ,  TR,  are  parallelograms  whose  opposite  sides  are 
equal,  IE  =  CQ,  and  TA  =  CR;  also  QE  =  CI,  and  RA=- 
CT. 

Therefore  (VII)  CQ  :  CR:  :QE-^  :  RA^ 

q.  E.  D. 


24  OF  THE  PARABOLA. 

Cor.  1. — This  is  true  of  ordinates  on  either  side  of  the  di- 
ameter, because  every  diameter  bisects  all  lines  in  the 
section  which  are  parrallel  to  the  tangent  at  its  vertex. 
(5  Cor.  Prop.  IX.)  Hence  EV  and  AX  are  double  or- 
dinates :  and,  conversely,  as  all  double  ordinates  are  par- 
allel to  each  other,  and  are  bisected;  any  straight  line 
which  bisects  two  parallel  straight  lines  in  a  Parabola,  is  a 
.diameter. 
'^'  "'*  Hence  a  segment  of  a  Parabola  being  given  as  ONI,  it  is 
easy  to  find  its  diameter.  For  draw  any  other  line  as  C  L 
parallel  to  10,  bisect  them  both,  in  M  and  D,  the  line  MD, 
produced  until  it  meets  the  curve,  is  the  diameter  of  the 
segment  of  which  I  O  is  a  double  ordinate. 

CoR.  2. —  Hence, (as  in  Cor.  2.  Prop.  I.)  the  parameter  of 
any  diameter  being  a  third  proportional  to  its  absciss  and 
ordinate,  is  a  constant  quantity,  and  the  rectangle  of  any 
absciss,  and  the  parameter  of  that  diameter,  is  equal  to  the 
square  of  the  ordinate,  or  in  otiier  words  it  is  equal  to  the 
rectangle  of  the  parts  into  which  the  diameter  divides  the 
double  ordinate.  (See  Cor.  1.) 
J'?' 17  ThusCQ  P'=QE2=EQ  QV, 
And    CR.P"=RA-^==AR.  RX. 


PROPOSITION  xr. 

Any  straight  line,  in  a  Parabola,  terminated  by  the 
curve,  cuts  off  from  all  the  diameters,  which  it  intersects, 
parts  which  have  the  same  ratio  to  each  other^  as  the  rect- 
angles   of  the  segments  into  which  they  divide  the   line. 

Fig.  18      Draw  another   line  as  KI,  parallel  to  EG,  the  line  HF 
which  bisects  them  both,  is  a  diameter,  (Cor.  I,)  and  the 
halves  of  these  lines,  as  HI  and  FG  are  ordinates  to   thrs 
diameter,  of  which  VH  and  VF  are  the  abscisses; 
Then        VH  :  GL:  :KH  .  HI  :  KL  .  LI. 


and    '' 

VF 

therelore 

FH 

or 

FH 

.:. 

FH 

and 

FH 

or 

VH 

OF  THE  PARABOLA  c^^ 

For(XCor.2,^VH  .  F=:HI% 
.F'=FG% 

.P'=:HP-FG% 

•  t"  =  KL  .  LI,  5  i^ror 

.  P'  :  VH  .  P:  :KL  .  LI  :  HP=KH.5DCon 

HI.  verse. 

:  VH::KL.LI  :  KH  .  HI. 
:FH(oi  GL)  ::KH.HI  :  KL  .  LL 

Q.  E.  D, 

Cor.  1 . — As  the  demonstration  is  equally  applicable  to 
all  lines,  which  are  parallel  to  Kl,  the  proposition  may  be 
stated  universally  — The  parts  of  all  diameters  cut  off  by 
lines  parallel  to  each  other  in  the  Parabola,  are  to  each 
other,  as  the  rectangles  of  the  segments  into  which  the  di- 
ameters divide  the  parallel  lines. 

That  is  VF  ;  GL::EF.FG  :  KL  .LI. 

The  proposition  just  stated  evidently  includes  Prop.  X 
and  XI  (since  the  rectangle  of  the  equal  parts  of  a  doub- 
le ordinate,  is  the  same  as  the  square  of  the  ordinate) 
and  therefore  also  I  and  II,  which  are  included  in 
them. 

CoR.     2. —  As    the    proposition   is    applicable    to    any 
parallel  lines,  as  HL,  MN,  and   DE,  FG  ;   therefore  by  Fig.  22. 
equality  of  ratios,  the  rectangles  of  corresponding  segments 
into  which  parallel  lines,  in  a  Parabola  divide  each  other, 
have  to  each  other,  constantly  the  same  ratio; 

Viz. — The  ratio  of  the  parts  of  the  diameters  intersect- 
ed by  them. 

That  is  HR  .  RL  :  HI  .  IL::DR  .  RE  :  Fl  .  IG. 
and       DX.XE  :  FO  .  OG :  :MX  .  XN  :  MO.  ON. 


CoR.  3. — The  1  and  2  Corrollaries,are  applicable  to  the 
segments  of  parallel  lines,  intercepting  a  diameter,  or  in- 
tersecting each  other  produced,  without  the  section. 

ThatisKR.RH   :KE.EH::PR  .RN    :  EL^.  rig;.  2j. 

and     PD  .  DN  :  FL^ :  iCD^  :  CF«.     (See  Prop.  3.) 


26  OF  THE  PARABOLA. 

Fig.  18.  (^oR  4. — It  is  also  true  universally,  that  the  rectangle 
of  the  paranneter  of  any  diameter,  and  the  part  of  anv  oth- 
er diameter  cut  off  by  a  double  ordinate  to  the  former,  is 
equal  to  the  rectangle  of  the  sef];ments  into  which  (hat  dou- 
ble ordinate  is  divided  by  the  second  diameter. 

That  is  P  .  GL^KL  .  LI. 

See  Prop.  XI  Cor.  2.  Also  I  Cor  2  and  II  Cor  1 . 

Cor.  5.     As  the  parameter  of  any  diameter, 
Is  to  the  sum  of  any  two  ordinates. 
So  is  the  difference  of  those  ordinates 
To  the  difference  of  their  abscisses. 

That  is  P  :  KL::LI  :FH=GL. 
(See  II  Cor.  2.) 


PROPOSITION  XII. 

If  there  be  a  tangent  to  the  curve,  and  any  line  be  drawn 
from  the  point  of  contact  to  meet  the  curve  in  some  other 
place,  and  if  any  diameter  intersected  by  this  line,  be  pro- 
duced to  meet  the  tangent,  then  shall  the  curve  divide  the 
diameter  in  the  same  ratio  in  which  the  diameter  divide? 
the  line. 

Fig.  19      That  is    ]E  :Ek::Ck  :  yfcP, 

6.4&22.     For  (VI)  IE 

6.  4.k  i.Also,sim.tri.IA;  : 

5.  3'2.     Therefore  IE 

5.A.&17.        And       IE 

Q.  E.  D. 

Cor.   l.--WhenCA;=A:P,   then  IE=EA:. 

Fig.  24.  Cor.  2. — If  TM  be  the  diameter  of  which  CL  is  a  dou- 
ble ordinate,  then  CD=DL  and  DN=NT,  and  as  the  de- 
monstration is  equally  applicable  to  a  tangent  to  either  end 
of  the  double  ordinate,  the  two  tangents  CT,  and  LT,  meet 
the  diameter  in  the  same  point. 

Cor.  3.  And,  conversely,  if  any  two  tangents  as  CT, 
LT  intersect   each   other,    the  diameter  which     passes 


TP: 

:CP  :  CT'y.Ck^  :  C?k 

TP: 

:Ck  :  CP   :\Ck^  :  Ck  ,  CP. 

U: 

\Ck.CP    :CP2  ::CA::CP, 

EA:: 

:CA::  A:P. 

OF  THE  PARABOLA.  27 

through  the  point  of  intersection,  bisects  the  line  which 
connects  the  two  points  of  contact,  or  this  line  is  a  double 
ordinate  to  that  diameter. 

Also,  if  the  line  which  is  drawn  through  the  intersec- 
tion of  the  tangents,  bisects  the  line  in  the  circle,  the 
former  is  the  diameter  of  which  the  latter  is  a  double 
ordinate. 

ScHOL. — Hence,  (Cor.  1,)  a  tangent  may  be  drawn  from 
any  point  without  the  curve,  as  T.  Draw  TM  parallel  to 
the  axis,  take  ND  =  NT,  draw  an  ordinate  from  D,  which 
will  cut  the  curve  in  C  the  point  of  contact.  The  ordinate 
DC  is  to  be  drawn  parallel  to  NR  which  may  be  drawn  by 
Schol.  Prop.  IV. 


PROPOSITION  XIII. 

If  from  any  point  of  the  curve  a  tangent  and  ordinate  be 
drawn  to  any  diameter,  and  another  ordinate  be  produced 
to  meet  the  tangent,  then 

As  the  difference  of  the  ordinates, 

Is  to  the  difference  added  to  the  external  part, 

So  is  double  the  first  ordinate, 

To  the  sum  of  the  ordinates. 

That  is  HO  :  HP:  :Hn  :  HI.  Fig.  24. 

For(def.  19,)  P'  :  DC::DC  :  DN, 

and  P'  :2DC::DC  :2DN=DT;  5.4. 

But  Sim.  tri.  DC  :  DT:  .HP  :  HC,  .   . 

therefore      HP  :  HC :   P'  :  2DC,  5.  li. 

or  HP  :  P' :  :HC  :  (2DC=)CL;  s.  i6. 

again  II  Cor.  HO  :P'::HC  :  HI, 
therefore      HO  :  HP:  :(CL=)Hn  :  HI.  5.  23. 

Q.  E.  D, 


Cor. — Hence,  (by  division,  composition,  inversion 
alternation  &c.)  PH  is  a  mean  proportional  between  PO 
and  PI,  or  PO  :PH::PH  :  PI. 

ScHOL.-Hence  a  tangant  can  easilybe  drawn  to  the  curve, 
from  any  point  without  it,  as  P.  Draw  PHI  intersecting 
the  curve  in  O  and  I,  and  take  PH  a  mean  proportional 
between  PO  and  PI  ;  then  draw  HC  parallel  to  the  axis, 
and  C  will  be  the  point  of  contact. 


2S  OF  THE  PARABOLA. 


PROPOSITION  XIV, 


The  distance  from  the  Focus  to  the  vertex  of  any  diame- 
ter, is  equal  to  one  C)urth  of  its  Parameter. 
Kig.  2a         That  is,  P=4FC;  or  {P=FC. 

For  Hiaw  the  ordinate  MA  parallel  to  the  tangent  CT, 
as  also  CD,  perpendicular  to  the  axis  AD,  and  FH  perpen- 
dicular to  the  tangent  CT; 

The  (IV)     AD  =  AT==CM, 

and  P.  AD  =  CD^;alsoP.CM=MA2 

5.  14.  therefore      P\   AD  :  P  .  CM: :  CD=  :  MA^, 

5.  23.  and  since     CM  =  AD.'.P  :  F'y.CD'^  :  MA2=:CT% 

.   .     But  by  sim.  tri.    FH  :  FT:  :CD  :  CT, 
6.22.         and  FH    :FT2::CDMCTS 

3.  11.      therefore         P  :  P':  :FH2  ♦  pT^. 

bit  (VIll.  Cor.  5.)  FH^  =FA  .  FT, 

,  .  .-.  P:P'::FA.FT  :  FT^ ; 

I]:  and  P:P'::FA:FT=FC; 

But  (III)         P  =  4FA, 
therefore  P'=4FT,  or  FC  =  iP. 

Q.  £.  D. 

Cor.  1 . — If  from  each  of  the  vertices  of  two  diameters 
an  ordinate  be  drawn  to  the  other  diameter,  the  two  ab- 
scisses will  be  equal  to  each  other.     For  AD=AT  =  CM. 

Cor.  2. — The  distance  from  the  vertex  of  any  diameter 
to  the  ordinate  passing  through  the  Focus,  is  equal  lo  the 
distance  from  the  vertex  to  the  Focus;  and  is  therefore 
equal  to  ^  its  parameter.     That  is.  CN  =  FT=FC. 

Cor.  3 — Of  any  diameter,  that  double  ordinate,  which 
passes  through  the  Focus,  is  equal  to  the  parameter  of 
the  diameter. 

For  (def.  19.)  CN  :  NK:  :NK  :  P', 
But  (Cor.  2  )  CN=iP'.-.CN  .  P'  =  iF  .  P'=iP'S 


.-,.  c 


s.  1 


^',7     "         (19)      CN  .  P'=NK2.-.NK2  =  iP'2; 
"2.  8.'  or         P'2=4NKS  and  P'=2NK. 


OF  THE  PARABOLA.  29 

Cor.  4. — Hence,  and  from  (Cor.  3,   and  VII.  Cor.  4.) 
!  It  appears,  that  if  the  directrix  HTB   be  drawn,  and  any  Fig-  15. 
I  lines  HE,  HE  parallel  to  the  axis ;  then   every   parallel  HE 
I  will  be  equal  to  EF,  or  |  of  the  parameter  of  the  diameter 
i  to  the  point  E.     It  is  also  equal  to  CN  the  absciss    of  the  Fig.  23. 
!  Focal  ordinate. 

'  Cor.  5. — From  the  fourth  step  of  the  demonstration,  it 
appears  that  the  parameters  of  different  diameters,  are  as 
the  squares  of  their  ordinates,  which  are  equally  distant 
from  their  vertices. 

Cor.  6. — The  parameter  of  any  diameter  equals  the  pa- 
rameter of  the  axis,  added  to  four  times  the  absciss  of  an 
ordinate  drawn  from  the  vertex  of  that  diameter,  to  the 
axis. 

ThatisP'=P-h4AD. 

For  by  the  last  step  of  the  demonstration, 
P'  =  4FT=4FA  +  4AT=4FA+4AD. 
tnd     P  =  4FA.  .P'=P  +4AD. 

Cor.  7. — Also  any  two  straight  lines  which  intersect 
each  other  in  the  Focus  are  to  each  other  as  the  rectangles 
of  the  segments  into  which  they  are  divided. 

That  is  CG  :KE::CF.FG  ;  KF  .  FE. 

For  by  Cor.  3.  these  lines  are  the  parameters  of  those 
diameters  to  which  they  are  double  ordinates. 
Hence  by  (II  Cor.  1.)  P'  or  CG  .  AF  =  CF  .  FG, 

and   "  P'orKE.  AF-=KF  .  FE, 

therefore  CG  .  AF  :  KE  .   AF  :  :CF  .  FG  :  KF  .  FE.    5.  i^ 
and  CG  :  KE::CF  .  FG  :  KF  .  FE. 

And  generally;  if  two  straight  lines  intersect  each  other,    6.  i. 
i/i  any  point  within  the  section,  the  rectangles  of  their  seg- 
ments are  to  each  other  as  the  parameters  of  those  diame- 
ters to  which  they  are  double  ordinates. 

For  CG  and  KE  (Cor.  3.)  are  those  parameters, 
aDd(VinCor.2.)CF.FG  :  c/./^::KF.FE  i  Kf ,  fE, 
CG:KE::cf.fg:Kf.fE. 
5 


30  OF  THE  PARABOLA. 

Cor.  8.— The   subtangent    PT=NK=2CN=iP'  the 
parameter  of  the  diameter  CN. 
6.  8.     For.TD  :  TC::TC  :  TP, 

but     rD=2AD=2CM&iTC  =  AM, 
.-.    2CM  :  AM:  :AM  :  TP.  .2CM  .  TP=AMs 
butCM.  P'=AMa.  .2CM  :  TP(=2TP  .  CM)=P'CM, 
.i.     2TP=ForTP=iP=NK=2CN. 


PROPOSITION  XV. 

If  there  be  two  tangents  to  the  curve,  intersecting  each 
other,  and  if  from  the  points  of  contact,  lines  be  drawn  to 
the  focus,  these  lines  make  an  angle  which  is  double  to 
that  made  by  the  tangents. 
Fig.  25.      Thai  is,  the  angle  DFC=2DTC. 

Draw  the  axis,NMH,  produce  DT  ;  to  cut  the  axis  in  N, 
Then  (VIII,  Cor.  3.)  angle,  DFH=2DNH; 

And    '*     CFH=2CMH  ; 
Therefore  "     DFC=2DNH+2CMH=2TNM-|-2TMN; 
That  is,    "     DFC=2DTM=2DTC. 

Q.  E.  D, 

Cor. — Hence  D/c,  is  a  right  angle, 

For     DFH-{-cFH=2  right  angles, 
.i.      Dtc=a  right  angle. 


OF  THE  ELLIPSE. 


PROPOSITION  I. 

The  squares  of  any  two  ordinates  to  the  axis,  are  to 
each  other,  as  the  rectangles  of  their  abscisses. 

Let  RVB  be  a  triangular  section  through  the  axis  of  the  ^'S-  ^■ 
cone  (3),  AGIB,  another  section  perpendicular  to  the  for- 
mer, forming  an  Ellipse  ;  AB,  the  mutual  intersection  of  the 
two,  will  be  the  transverse  axis  of  the  Ellipse  (11);  let 
MIN,  KGL  be  circular  sections  parallel  to  the  base  (4); 
FG,  HI,  the  mutual  intersections  of  these  planes  with  that 
of  the  Ellipse,  will  be  ordinates,  both  in  the  Ellipse,  and 
in  the  circles,  being  perpendicular,  both  to  MN  and  KE, 
and  to  AB. 


Then     FG^  :  HI^ : :  AF  .  FB  :  AH  .  HB. 

Forsim.tri.  AF  :  AH::FL  :  HN, 

And  FB  :  HB::KF  :MH; 

Therect.  .'.  AF  .  FB  :  AH  .  HB::KF.FL  :  MH  .  HN,   6.  C. 

But  KF  .  FL=FG2,  andMH.HN=Hl%  ^__. 

Therefore  AF  .  FB  :  AH  .  HB:  iFG^  :  HI^. 

Q.  E.  D.  5.  7. 


ScHOL. — If  then  a  straight  line,  placed  perpendicularly 
upon  another  finite  straight  line,  were  to  move  parallel  to 
■itself,  and  to  vary  in  its  length  continually,  so  that  its 
squares  should  always  have  the  same  ratio  to  each  other, 
as  the  rectangles  of  the  parts  into  which  it  divides  the  oth- 
er line,  the  figure  formed  by  the  moving  line,  would  be 
the  portion  of  an  Ellipse  between  the  axis  and  curve. 


32  OF  THE  ELLIPSE. 

Cor.  1. — As  the  proposition  and  demonstration,  are 
equally  applicable  to  ordinates  on  opposite  sides  of  the 
axes,  having  the  same  abscisses,  those  ordinates  are  equal 
or  GS,  is  double  of  FG.  Hence  it  is  called  a  double  or- 
dinate. 

CoR.  2. — The  squares  of  the  ordinates  are  the  same 
as  the  rectangles  of  the  equal  segments  of  the  double  or- 
dinates, therefore,  the  proposition  may  be  thus  expressed. 

The  rectangles  of  the  segments  of  the  double  ordinates,  art 
to  each  other  as  the  rectangles  of  the  abscisses,  or  segments 
into  which  they  divide  the  transverse. 

Cor.  3. — Ordinates  at  equal  distances  from  either  ver- 
tex, are  equal  ;  since  the  abscisses  and  consequently  the 
rectangles  of  the  abscisses  are  equal.  And  conversely,  if 
the  ordinates  are  equal,  their  distances  from  the  vertices 
are  equal. 

Cor.  4. — Hence  the  Foci  are  equally  distant  from  eith- 
er vertex,  for  their  ordinates  are  equal,  being  each  equal 
to  the  semi-parameter,  (20),  their  distances  from  the  cen- 
tre are  therefore  equal. 

Cor.  5. — Hence  also  every  diameter  is  bisected  in  the 

centre.     For  since  at  equal  distances  from  the  centre,  the 

Fi?.  t4ordinates  are  equal,  the  third  sides  and  the  remaining  an- 

l    14.  ^^^^  ^'  ^^^  right  angled    triangles,  Cr/e,   C/?g  are  equal, 

therefore  ecg,  is  a  right  line  bisected  in  the  centre. 

And  conversely,  the  ordinates  from  the  ends  of  any  diam- 
eter upon  the  transverse  axis  are  equally  distant  from  the 
centre. 

CoR.  6. — Hence  also,  the  whole  Ellipse,  is  by  the 
transverse  axis,  divided  into  tw^o  equal  parts,  which  if  pla- 
ced one  upon  the  other  will  coincide  in  every  respect. 
For  if  they  should  not  coincide  in  any  part,  the  or- 
dinate in  one,  at  that  point,  would  be  unequal  to  the  cor- 
responding ordinate  in  the  other. 

CoR.  7. — For  similar  reasons,  the  two  parts  of  the  tan- 
gent at  the  vertex,  would  coincide  ;  that  is,  the  tangent  at 
the  vertex  makes  equal  angles  with  the  axis,  on  each  side 
of  it.     It  is  therefore  perpendicular  to  the  axis. 


OF  THE  ELLIPSE.  33 

Cor*  8.  If  a  circular  section  as  P  C  O.  pass  through  the    ^'S-  1 
centre  of  the  Ellipse,  the  semi-conjugate  Ca,  is  a   mean 
proportional  between  OC  and  CP,  the  parts  of  the  diam- 
eter of  the  circular  section  ;  for  in  this  circle  Ca  is  an  or- 
dinate to  this  diameter. 

Also  the  whole  conjugate  axis,  is  a  mean   proportional 
between  BR  and  AQ  ,  the  diameters  of  the  circular  sec 
tions,  passing  through  the  two  vertices  of  the  Ellipse. 

For  as  AB=r2AC,.-.BR=2CP,  and  QA=20C, 

therefore  BR  :      ab    '.'.ab  :  AQ.  5,  16. 

ScuoL. — It  has  been  already  remarked,  that  if  the  cut- 
ting plane  be  supposed  to  revolve  about  A,  the  axis  AB  will 
become  more  extended,  the  abscisses  FB  and  HB,  will  ap- 
proach to  a  ratio  of  equality,  and  the  section  itself  will  ap- 
proach indefinitely  near  to  a  Parabola  ;  and  when  it  be- 
comes parallel  to  the  side  of  the  cone,  it  will  be  a  Parabo- 
la (o).  \{  then  the  indefinite  abscisses  are  considered 
equal,  the  preceding  proposition  will  become, 

AF  :  AH::FG2::Hl*  e.  i. 

the  same  as  the  first  proposition  of  the  Parabola. 

Again,  if  the  cutting  plane  revolve  in  the  opposite  di- 
rection, it  will  come  into  coincidence  with  the  circular 
section  A  Q.  parallel  to  the  base  of  the  cone.  As  the  ratio 
of  the  ordinate?  and  abscisses  remains  the  same  during 
the  revolution,  we  conlude,  that  it  will  be  found  also  in 
the  circle.  This  property  of  the  circle  is  demonstrated 
by  Euclid  ;  compare  31  Prop,  in  the  3d  Book  with  the  8th 
in  the  6th  Book,  from  which  it  will  appear,  that  if  ordi- 
nates.  or  perpendiculars,  be  drawn  to  the  diameter  of  a 
circle,  their  squares  are  respectively  equal  to  the  rectan- 
gles of  their  abscisses.  The  squares  therefore  have  to 
each  other,  the  same  ratio  which  the  rectangles  have,  and 
its  being  the  ratio  of  equaliti^  is  what  distinguishes  the  cir- 
cle from  the  Ellipse. 

Note. — When  the  ronrcrac  of  propositions  are  stated  in  this  work,  the 
demonstrations  are  not  added,  because  they  are  easily  made  out  by  a  rcdur- 
Ho  ad  absurdum,  by  which  it  is  proved  that  to  deny  the  convei-seof  the 
proposition  is  in  effect  to  deny  the  proposition  itself  of  which  it  is  the 
converse.     (See  Eucl.  XVIII  and  XIX,  XXIV  and  XXV,  &c.) 


34  OF  THE  ELLIPSE. 


PROPOSITION   11. 


As  the  square  of  the  Transverse  Axis, 
Is  to  the  square  of  the  conjugate  Axis, 
So  is  the  rectangle  of  the  abscisses  of  the  Transverse. 
To  the  square  of  their  ordinates. 
Fig.  2.      That  is  AB»  :  a6»  : :  AD  .  DB  :  DE^. 

For,  (1)     AC  .  CD  :  AD  .  DB:  :Ca^  :  DES 
But  if  C  be  the  centre,   AC  .  CB=AC2,  and   Ca  is  the 

semi-conjugate, 
Therefore  AC^  :  AD  .  DB:  :Ca^  :  DES 
and       AC2  :  Ca^ : :  AD  .  DB  :  DE% 
^  15.         or      AB2  :  ab^ : : AD  .  DB  :  DE^ 

Q.  £.  D. 

2-  5.    Cor.  1. —Since  the  rectangle  AD  .  DB^CA^  -CDS 

Therefore     AC^  :  aC^ :  :CA»  -CD^  :  DE^; 
or  AB^  :  ab'  :  :CA«— CD^  :  DE^. 

Cor.  2.— (19)AB    lab    lab    :  Parameter=P, 

Cor  2.        a'so        AB    :  P    ::AB2  :  ab^, 
5.11.  .'.  AB    :P    ::AD.DB:DES 

or  The  Transverse  axis, 

Is  to  its  Parameter, 
As  the  rectangle  of  its  abscisses, 
To  the  square  of  their  ordinate. 

Cor.  3. — If  different  Ellipses  have  the  same  Transverse 
axis  the  corresponding  ordinates  in  each,  will  be  to  each 
other  as  their  Conjugate  axes. 

For  as  the  tirst  and  third  terms  in  the  proposition, 
will  then  be  the  same,  the  second  will  vary  as  the  fourth  •, 

|';&vf;        or       AC3  :  ad  .  DB:  :C«^  :  DEs 

and        AC»  :  AD  .  DB-.Ca^  :  DeS 

Ca^  :  Ca'2   ::DE2  :  De% 
and        Ca    :  Ca      ::DE    :  D«. 

Fig.  15.      CoR.  4. — Let  AFGB  be  the  rectangle  of  the  Transverse 
axis,  and  Parameter  BG,  then  DHGB  is  the  rectangle  of 


OF  THE  ELLIPSE.  35 

the  absciss  DB  into  the  Parameter,  and  DTFB  or  the  rec- 
tangle Dl  .  DB  is  equal  to  the  square  of  DE. 

For  (CoR.  2.)  AB  :  BG::AD  .DB  :  DES 
andSim.  tri.     Af^  :  bG: :  AD  :  Dl ! :  AD  .  DB  :  DI  .DB,    6.  >• 
and  AI .  .  DB  :  DE=» : :  AD  .  DB  :  DI  .  DB  ; 

.:.  DE2=D1.DB. 

ScHOL. — The  square  of  the  ordinate  DE,  therefore 
(which  equals  DIFB)  is  less  than  the  rectangle  of  the  ab- 
sciss DB  into  the  Parameter,  or  DHGB,  by  the  rectangle 
IHGF  similar  and  similarly  situated  to  the  whole  rectan- 
gle PB  It  was  on  account  of  this  deficiency  in  the  square 
of  the  ordinate,  compared  with  the  rectangle  of  the  ab- 
sciss into  the  Parameter,  that  Apollonius  named  this  sec- 
tion the  Ellipse,  The  rectangle  ABGP  or  AB  .  P  was 
called  the  Figure  o(  the  section  ;  AB  and  BG  its  latera, 
and  the  perpendicular  BG  the  Latus  rectum. 


PROPOSITION   III. 

As  the  square  of  the  Conjugate  axis, 

To  the  square  of  the  Transverse, 

So  is  the  rectangle  of  the  abscisses  of  the  Conjugate, 

To  the  square  of  their  ordinate. 

That  is  fl6»  :  AB^ :  :ad  .  db  :  dE^. 

Draw  ED  an  ordinate  to  the  Transverse  AB,  Fig.  2. 

Then  (IICor.  1)  AC^  :  aC^r.-CA^-CD^  :  DE^ ; 

But  CD^  =t/E%  and  DE^  =C(/^ 

Therefore       AC^  :  aC^  wCA'—dE^  J  CJS 
and  AC^  :  CA2  -  dE^ :  laC^  :  Cd^, 

„  AC2  :  c/E2 :  :aC2  .  ^q^  -Cd\  5.  E. 

„  AC^  XaC^y.dE^    XaC^-Cd\ 

aC^  :  kC^WaC^-Cd^  :  dE'  ; 
But  flC^  -  C(/^  =ad  .  db,  2.  s.Cor. 

Therefore       aC^  :  AC'l'.ad^  db  t  dE''. 

Q.  £.  D. 


36  OP  THE  ELLIPSE. 

Cor.  I. — From  this  Proposition  it  appears  thaUhe  Con- 
jugate axis,  and  the  rectangles  of  its  abscisses,  have  the 
same  relations  to  their  orditiates  and  to  the  Transverse, 
which  is  its  conjugate  (16),  as  those  which  have  been  de- 
monstrated concerning  the  Transverse  axis. 

Hence  1st  the  Conjugate  axis,  also,  bisects  all  its  doub- 
le ordinates. 

2.  Ordinates  at  equal  distances  from  its  vertices,  or 
from  the  centre,  are  eijual. 

3.  The  Conjugate  axis  divides  the  Ellipse  into  two  equal 
and  similar  parts. 

4.  The  tangents'at  its  vertices  are  perpendicular  to  it. 

5.  Ordinates  from  the  opposite  extremities  of  any  di- 
ameter are  equal  and  equally  distant  from  the  centre. 

6.  The  square  of  the  Conjugate  axis,  is  to  the  square  of 
the  Transverse,  as  the  difference  between  the  squares  of 
the  semi-conjugate,  and  the  distance  from  the  centre  to 
any  ordinate,  is  to  the  square  of  that  ordinate. 

7.  If  two  Ellipses  have  the  same  Conjugate  axis,  the 
corresponding  ordinates  to  this  axis,  will  be  to  each  other 
as  their  Transverse  axes. 

8.  The  Conjugate  axis  is  to  its  Parameter,  as  the  rect- 
angle of  its  abscisses,  to  the  square  of  their  ordinate. 

9.  The  square  of  any  ordinate  is  less  than  the  rectan- 
gle of  its  absciss  ana  the  Parameter  of  the  Conjugate,  by  a 
rectangle  similar  and  similarly  situated  to  the  rectangle  of 
the  Conjugate  and  its  Parameter.     (See  Prop,  i  and  II.) 

10.  The  squares  of  any  two  ordinates  areas  the  rec- 
tangles of  their  abscisses  ;  or  in  other  words,  the  rectangles 
of  the  segments  of  its  double  ordinates  are  as  the  rectan- 
gles of  the  segments  into  which  they  divide  the  Conjugate 
axis. 


OF  THE  ELLIPSE.  37 

ScHOL.  From  Prop.  Ill  compared  with  Prop.  II,  it  is 
evident,  that  if  two  unequal  straight  lines  bisect  each  oth- 
er at  right  angles,  and  either  of  them  move  parallel  to  it- 
self on  the  other,  varying  in  its  length  contii»ually,  so 
that  the  rectangles  of  its  parts  have  always  the  same  ra- 
tio to  the  rectangles  of  the  parts  into  which  it  divides  the 
other  line, — the  figure  formed  by  the  moving  line  will  be 
an  Ellipse. 

Cor  2. — If  two  circles  be  described  on  the  two  axes  of 
an  Ellipse  as  diameters,  the  one  being  inscribed  within  the 
Ellipse,  and  the  other  circumscribed  about  it,  then  any 
ordinate  in  either  circle  is  to  the  corresponding  ordinate 
in  the  Ellipse  as  the  axis  of  this  ordinate  is  to  the  other 
axis. 

That  is,  CA  :  Ca::DG  :  DE. 

and      CaiCA::  dg  :  dE. 

For  AD    .  DB  =DGS 

Therefore  (II)       CA^  :  Ca^  :  iDG^  :  DE^  ; 
Or  CA    :Ca::DG    :DE. 

In  the  same  manner.  Ca  I  CA'.'.dg  t  dE, 

DGtDE::dE::dg. 

Or  the  corresponding  ordinates,  in  the  circlei  and  El 
lipse  are  reciprocally  proportional. 


PROPOSITION  IV. 

The  square  of  either  axis,  is  the  square  of  the  other,  as 
the  rectangle  of  the  segments  of  a  line  in  the  Ellipse  paral- 
lel to  the  former,  is  to  the  rectangle  of  the  corresponding 
segments  of  a  line  parallel  to  the  latter. 

Fig.  3. 


That  is     eh 

.  hg  :  Eh  . 

AG::AB2  lab^ 

Or       Eh 

,  hG  :  eh  . 

hg'.'.ab^  :  AB\ 

38  OF  THE  ELLIPSE 

5.   E. 


For  (I)  ED»  :  eN^ : :  AD  .  DB  :  AN  .  NB, 
and  ED^  :  ED^— eN^ :  lAD.DB  :  AD.DB-AN.NB. 


But  ED^-e]^^=ED'-hD'=-Eh  .  AG, 


And  AD.DB-AN.NB=BD.DN+NA-BD+DN.NA; 


2. 1.    And  since,  BD  .  DN-hNA  =  BD  .  DN-f-BD  .  NA, 


And  BD  -f-  DN  .  NA=BD  .  NA+DN  .  NA, 

Therefore,  AD.DB-- AN  .  NB=DN  .  BD-NA  .  DN 

=  Dw  .  DN  =  %    .  eh  ; 

ED=  :  Eh  .  AG:: AD  .  DB::eh  .  hg, 
And  ED^  :  AD.  DB::EA  ,  hG  :  eh    .  hg. 

But    (II.)      ED2  :  AD  .  DB::«62  :  AB^, 
11.  .:.  Eh  .  hG  t  eh  ,  hgWab'-  :  AB^. 

Q.  E,  D. 

Cor.  1 . — Hence  by  equality  of  ratios,  eh  .  hg  :  ek  ,kg'.\ 
Eh  .  hG  :  M'  .  kl.  Or,  universally,  the  rectangles  of  the 
correspondiijg  segments,  of  lines  parallel  to  the  two  axes, 
mutually  intersecting  each  other  in  the  Ellipse,  are  to 
each  other  always  in  the  same  ratio;  namely,  the  ratio  of 
the  squares  of  the  two  axes. 

ScHOL. — As  the  square  of  either  semi-axis,  or  of  an  or- 
dinate, is  the  same  as  the  rectangle  of  the  two  halves  of  the 
axis  or  double-ordinate.  This  proposition  evidently  in- 
cludes Prop.  II  and  III,  as  it  demonstrates,  that  the  rela- 
tions there  shown  to  exist  between  the  two  axes,  extends 
to  all  lines  in  the  section  parallel  to  them. 

For       AC2  :  ac^::  AD  .  DB  :  DE^    is  the  same  as 
AC    .  CB  :  aC  .  C6: :  AD  .  DB  :  ED  .  DG 
which    may  be   considered    as    included   in   this    propo- 
sition. 

If  the  cutting  plane  revolve  until  the  section  becomes  a| 
circle,  we  should  infer,  that  the  rectangles  of  lines,  mu- 
tually intersecting  each  other  at  right  angles,  in  the  cir- 
cle, have  always  the  same  ratio.  This  is  demonstrated 
by  Euclid,  (HI  Book,  35  prop.  See  Schol.  to  Prop,  I.) 
and  also,  that  it  is  a  ratio  of  eiquality. 


OF  THE  ELLIPSE.  39 

If  the  plane  be  supposed  to  revolve  in  the  opposite  di- 
rection until  the  section  becomes  a  Parabola,  and  the  lon- 
ger segments  of  the  lines  parallel  to  the  transverse  axis  be 
considered  equal. 

Then  the  Prop.  AD  .  DB  :  eh  .  ^^^:  :ED  .  DG  :  EA  .  /iG 
becomes        AD  :  eA : :  ED  .  DG  :  E/i .  AG 
The  same  as  is  demonstrated  in  the  Parabola,  Prop.  II. 
Cor.  1. 


PROPOSITION.  V. 

If  lines  parallel  to  the  two  axes,  intersect  each  other, 
without  the  Ellipse,  the  rectangU:s  of  their  corresponding 
segments  are  to  each  other  as  the  squares  of  the  axes  to 
which  they  are  parallel. 

Fie.  4. 


That  is,  EH  , 

.   UG  :  eU  , 

Hgy.ab''  :  ABS 

or      eU  . 

Hg : EH  . 

hG::AB=  :  ab*. 

For  (I,)  e^^  :  ED^ : :  AN  .  NB  :  AD  .  DB, 
And    eN=  :  eN^-ED2::AN.NB  :  AN.NB-AD.DB. 

But    .N2-ED-^=HD— ED2=EH  .HG, 


Also  AN  .  NB  =  BN  .  AD-fND=BN  .  AD-f  BN  .  DN, 


And    AD  .  DB=AD  .  B    +ND=BN  .  AD-f  AD  .DN, 
Therefore  AN  .  NB  -  AD  .  DB=BN  .  DN  -  AD  .  DN, 


=DN  .  BN-AD=DN  .  BN-Br/ 
=DN.  ^d=DN  .  Dn. 

Therefore  (eN2::=)HD2  :  EH  .  HG:  :AN  .  NB  :  (DN . 
Dn  =  )eU   .   H^, 
And        HD2  :  AN  .  NB::EH  .  HG  :  eH   .  H^; 
But  (II.)    HD2(=eN0  :  AN  .  i\B::   b^  :  AB-', 
EH    .     HG     :   eU   .  Eg  Wah^  :  AB^ 

Q.  E,  D. 


40  OF  THE  ELLIPSE. 

Cor.  1.  — Since  this  proposition  is  equally  applicable 
to  all  lines  parallel  to  the  axes,  it  follows  by  equality  of 
ratios^  that,  the  rectangles  of  the  segments  of  all  lines 
parallel  to  the  axes,  intersecting  each  other  wtthout  the 
Ellipse,  are  to  each  other  always  in  the  same  ratio  ;  viz. 
that  of  the  squares  of  the  two  axes. 

That  is     EH  .  HG  :  eH  .  H^::KI  .  IL  :  el  .  ]g. 

CoR.  2. —  If  DH  be  supposed  to  move  towards  AP,  the 
rectangle  EH  .  HG  approaches  continually  to  an  equali- 
ty with  the  square  of  AP,  as  its  limitf  and  when  D  coin- 
cides with  A,   AP-  ma)  be  substituted  for  EH  .  HG. 
In  like  manner,  aS^  may  be  subitituted  for   eH  .  HG. 

That  is,  the  rectangles  of  the  segments  of  lines  parallel 
to  the  axes,  are  to  each  other,  as  the  squares  of  the  inter- 
cepted parts  of  the  tangents  to  the  vertices  of  the  axes. 

^'?-  3.      Cor.  3. -Since  (IV,)  Eh  .  hG   :  eh  .  hg  y.ab'  :  AB-% 

^'^'  '^'  and  from  this  EH  .  HG  :  eH .  H^  :  ab^  :  AB^, 

Therefore     Eh  .    hG  I  eh  ,  hg:  :EH  .  HG  :  eH  .  H^. 

ScHOL.  — When  the  Ellipse  becomes  a  circle,  the  same 
relations  exist  betwen  the  rectangles  of  the  corresponding 
segments,  without  the  figure,  and  also  between  these 
rectangles  and  the  squares  of  the  tangents,  as  is  demonstra- 
ted in  Euclid  3  Book  36  and  37,  but  it  is  then  a  ratio  of 
equality. 

Also  if  the  plane  revolve  in  the  contrary  direction,  un- 
til the  section  becomes  a  Parabola,  and  the  longer  seg- 
ments are  considered  equal,  then 

EH  .  HG  :  eH.H^y.Kl  .  IL  :  el  .  I^. 

becomes    EH  .  HG  :  eH:  :KI  .  IL  :  el  ; 
and       EH  .  HG  :  Kl  .  IL:>H  .el. 
As  demonstrated  of  the  Parabola  II,  Cor.    1. 

The  five  preceding;  propositions  may  all  be  embraced  id  ' 
the  (o\\o\\\n^,  gentrnl  proposition.     If  lines  parallel  to  one 
axis  rf  the  Ellipse  iutersect  the  other  axis,  or  lines  parallel 
to  it,  tithtr  withiji  or  without  the  section  ;  the  rectangles  of 


OF  THE  ELLIPSE.  41 

ihe  corresponding  segments  will  always  have  to  each  other, 
the  same  ratio. 


PROPOSITION  VI. 

If  a  tane^ent  and  ordinate  be  drawn  to  any  point  of  an 
Ellipse,  meeting  the  transverse  axis  produced,  the  »emi- 
traiisverse  is  a  mean  proportional  between  the  distances 
of  the  two  intersections  from  the  centre, 

That  is,  or  CM  .  CT=CA2  ;  CM  :  CA  :  CT,  are 
continued  proportionals.  Fig.  5. 

For  from  the  point  T  draw  any  other  line  TEH  to  cut 
the  curve  in  two  points  E  and  H  ;  from  which  let  fall 
the  perpendiculars  ED  and  HG  ,  bisect  DG  in  K, 

Then     (I)     AD  .  DB    :  AG     .GB::DE2:GH% 
And  bysim.  tri.  TD^  :  TG^* :  iDE^  :  GH^  ; 

Therefore  AD  .  DB  :  AG  .  GB  ::TD-^  :  TG^ 
But  DB=CB4-CD=AC+CD  =  AG+DC-CG=  iCK 

+  AG, 
AndGB=CB-CG=AC-CG=AD+DC-CG=2CK 
+  AD. 
.-.    AD  .  2CK-f  AD  .  AG  :  AG  .  2CK+AG  .  AD:: 
TD2 : TG% 
and  DG  .  2CK  :  (TG^  -TD^or)  DG  .  2TK: :  AD  .  2CK 

-f-AD.  AG  :TD%  ^^- Vl^^ 

or  2CK  :  2TK: :  AD  .  2CK-f  AD  .  AG  :  TD=^        *  ^ 

orAD.2CK  :  AD    2TK:  :AD  .2CK-f-AD .  AG  :  TD^;  5.  19. 
.-.  AD.2CK  :  AD.2TK::AD.  AG  :  TD2-AD.2TK, 
and    CK  :  TK : :  AD  .  AG  :  TD^  -  AD  .  2TK,  5.  a  &  is. 

and     CK:TC::AD  .  AG  :  TD^-AD  .  TD-f-TA ; 
or      CK  :  CT::AD  .AG  :  AT^. 

But  the  limit  of  AD  .  AG,  when  the  line  TH  comes  in- 
to the  position  ofTL,  is  AM^  (See  Parab.  Prop.  IV,)  and 
K,  then  coincides  with  M.  The  proposition  therefore 
becomes.  CM  :  CT: :  AM^  :  AT^  : 


42  Oi^  THE  ELLIPSE. 

That  is  CM  :  CT:  :CA-CM='  :  CT-CA«, 
5.19.       or     CM  :  CT::CIVP  +  CA^  :  CA2+CT^ 
and     CM  :MT::CM2+CA2  :  CT-^ -CM% 

or      CM  :  MT : :  CM^  +CA3  :  CT+CM  .  MT, 
or    CM^  :  CM  .  MT  : :  CM^  ^-CA^  :  CM  .  MT+ 
CT  .  MT ; 
Hence  CM^  :  CA-^ :  :CM  .  MT  :  CT  .  MT, 
i^onr^       3'^^    CM2  :  CA^XCM  :  CT, 
2  coa-''      •'•      CM  :  CA  : :  CA  :  CT. 

versely.  Q,  E.    D. 

The  converse  of  this  poposition,  may  be  very  concisely 
demonstrated  in  the  following  manner,  which  as  it  does 
not  depend  on  the  preceding  demonstration,  may  be  made 
the  principal  demonstration,  and  the  preceding  inferred 
as  the  converse  of  it. 


fig.  5.  On  AB  describe  the  circle  AGB,  produce  ML  to  G, 
draw  a  tangent  to  the  circle  at  G,  intersecting  the  axis 
produced  in    T,  join  CG,  then    CGT   is  a  right  angle, 

6.  8.  and  (CG=)CA,  is  a  mean  proportional  between  CM  and 
CT. 


Connect  T  and  L,  and  the  line  TL  is  a  tangent  to  the 
Ellipse  at  L.  If  not,  it  must  cut  it.  Let  it  be  supposed 
then  to  cut  it  at  L  and  e,  through  c  draw  the  ord.nate 
ed,  and  produce  it  to  cut  the  circle  in  g,  and  the  tangent 
in  r. 


Then,  sim.  triang.  TM  :  Td.'.ML  :  del 
and     "         "        TM  :  Tdr.MG  :  dr, 

Mh:MG::de:dr, 
but  (II,  Cor.  3.)  ML  :MG::de:  dg  (Cfl.  :  CA.), 

dr=dg  —  ihe  less  equals  the  greater, 
which  is  impossible.  Therefore  TL  does  not  cut  the 
Ellipse  in  L;  but  it  meets  it ;  it  is  therefore  a  tangent  to 
the  curve  in  L,  and  CT  is  a  third  proportional  to  CM 
and  CA. 

That   is   CM  :  CA  :    CT,  are   continued  proportion- 
als. 

Q.  E.  D. 


OF  THE  ELLIPSE.  43 

Cor.  1.— CM  :  CT::A]VP  :  ATS  that  is,  the  distances 
from  the  centre  to  the  two  intersections  of  the  tangent  and 
ordinate  with  the  axis,  are  as  the  squares  of  the  distances 
of  the  same  intersections  from  the  vertex. 


CoR.  2. — If  a  tangent  and  ordinate  be  drawn  from  anj 
point  in  the  Ellipse,  cutting  the  conjugate  axif;  produced, 
the  semi-conjugate  is  a  mean  proportional  between  the 
distances  of  the  two  intersections  from  the  centre. 

For  the  properties  of  the  transverse,  from  which  this 
proposition  has  been  deduced,  are  the  same   in  the  con-    '^' 
jugate.     See  Prop.  II  and  III. 

The  preceding  demonstrations  therefore,  are  applicable 
in  every  respect  to  the  semi-conjugate  axis. 

That  is  Cd  *,  Ca  :  C/,  are  continued  proportionals.  Fig.  6. 

CoR.  3. — As  the  demonstrations  are  applicable  to  a  tan-  Fig;.  7. 
gent  to  the  curve  at  either  extremity  of  the  double  ordin- 
ate, the  two  tangents  drawn  to  the  two  extremities  of  any 
double  ordinate  to  the  axis,  meet  the  axis  in  the  same 
point,  and  at  equal  angles. 

And,  conversely,  a  line  drawn  through  the  intersection 
of  two  such  tangents,  if  it  bisect  the  angle,  will  also  bisect 
the  double  ordinate  ;  and  if  it  bisect  the  double  ordinate, 
it  will  also  bisect  the  angle,  and  in  both  cases  it  is  the 
axis. 

Cor.  4. —  Also  KA:,  the  tangent  at  the  vertex  is  bisect- 
ed by  the  axis, 

For       TD:TA::DE  :  AK. 

and        TD  :  TA::DH  :  A^, 
therefore    DE  :  AK : :  DH  :  AAr ; 

but         DE=DH.-.AK=AA:. 
For  a  similar  reason,     NF=NG, 
And  as      NI  =NM  -.FI  =  MG. 


44  OF  THE  ELLIPSE. 

Cor.  3. — If  any  number  of  Ellipses  have  the  same 
transverse  axis,  and  an  ordinate  in  one  be  continued  ta 
intersect  the  curves  of  all,  the  tangents  from  all  the  points 
of  intersection,  will  meet  the  curve  in  the  same  point; 
for,  since  CD  and  CA,  the  two  first  of  the  continued  pro- 
portionals, remain  the  same  in  all,  the  third  CT  will  be 
the  same. 

It  continues  the  same,  if  the  Ellipse  becomes  a  circle, 
as  is  evident  from  the  first  steps  of  the  demonstration. 

CoR.  6. — Two  tangents  to  the  curve  at  the  extremities 
of  any  double  ordinate,  to  the  conjugate  axis,  cut  it  in  the 
same  point. 

CoR.  7. — Tf  any  number  of  Ellipses  have  the  same 
conjugate  axis,  the  tangents  drawn  to  each  at  the  inter- 
sections of  the  same  double  ordinate  produced,  will  cut 
th(   conjug    e  in  the  same  point. 

The  same  is  true  also  of  tangents  to  a  circle  described 
upon  the  conjugate  axis  as  its  diameter. 

^^S-  6.      Cor.  8. — The   following   list   of  proportionals    is   de- 
rived directly  from  this  proposition,  and  its  second  Corol- 
lary, viz. 
that    CT:CA::CA  :CD. 
And      Ct    I  Ca  y.Ca  :  Cd. 

Fig.  6.      (1.)  Then  CT-f  PA  :  CT-CA:  :CA+CD  :  CA-CD 
that  is  BT  :  AT::BP  :  AD. 
or     BT  is  harmonically  divided. 

In  the  same  manner  bt  :  at'.'.bd  *  ad, 

(2.)  Also  CT  :  CT-hCA::CA  :  CA+CD. 
that  is  CT  :  BT::CA  :  BD. 

And  Ct  :bt::Ca  I  bd. 

(3.)  Again  (2.)  CT-CA  :  BT-BD::CT  :  BT, 
That  is        AT  :  DT::CT:  BT. 

And,         at  :  di'.'.Ct  :  bt. 


OF  THE  ELLIPSE.  4^ 


(4.)  Therefore  sim.  tri.  AK  :  DE'.'.Ct  :  BH, 
and  "     "    TK  :  TE'.'.Tt  :  TH. 

also  ak  :  dE'.'.CT  :  bh, 

and  tk  :  tE'.uT  :  th. 


PROPOSITION  VII. 

The  rectangle  of  the  Focal  distances  from  the  vertices, 
is  equal  to  the  rectangle  of  one  fourth  the  parameter,  and 
the  transverse  axis. 

That  is  AF  .  FB==;iP  .  AB,  Tig.  8. 

For  (II  Cor.  -2.)  AF  .  FB  :  FE-^ : :  AB  :  P, 

therefore        AF  .  FB  :  FE^ : :  AB  .  AF  :  P  .  AF, 
and  AF  .  FB:  AB  .  AFiiFE^  :  P  .  AF ; 

But  (19)        FE  =  iP.-.FE2=iPS 

AF  .  FB  :  AF  .  AB:  :^P=^  :  P  .  AF,      g.s.Cor.s. 
FB  :  AB::iP  :  AF, 
and  FB  .  FA=iP  .  AB, 

Q.  E.  A 

Cor.   1.— FB  :  AB::iP  :  AF. 

ScHOL. — If,  as  in  the  Parabola,  FB  and  AB,  be  consid- 
ered equal,  then  :[P=AF  as  demonstrated  in  the  V  Prop, 
of  Parabola. 

CoR.  2.-SinceFB.FA=iP.AB,and  FB  is  less  than  AB, 

FA  is  greater  than  {P 
In  the  Parabola,  FA  =  iP. 

CoR.  3.— AF  .  FB=Ca\  or  the  rectangle  of  the 
Focal  distances,  equals  the  square  of  the  semi-conju* 
gate. 

For  since  (19.)  P  .  AB=ab^--'.i?  .  AB=aC2=:AF  .  FB. 
or  AF  :aC  ::aC  :  FB. 


46  OF  THE  ELLIPSE. 

Cor.  4.— AF.FB(=aC^)  =  ia6^  or  iAB.iP=AC.FE  ; 
or  AF  :  FE::AC  :FB. 

Cor.  3.— (VI  Cor.  1.)  aC  =  =C/  .  Cd=Ct  .  DE. 
but  (V(.  Cor.  7,  4.)C/  .  DE=AK  .  BH, 
.-.     AK  .  BH=aC2=CA  .  iP=AF  .  FB. 

CoR.  6. — Therefore  HFK,  is  a  right  angle, 
for  (Cor.  5)  AK   :   AF::FB  :  Bfl, 
.'.  the  triangles  AKF  and  BHF     -e  -infi''  r; 
-^  .-.    -  angle  BHF=AFK  and  BFH  =  AKF, 

.*.  also  HFK  is  a  right  angle. 

The  same  may  be  proved,  in  referenre  to  the  other 
Focus,  therefore  a  circle  described  on  HK,  as  a  diameter, 
will  pass  through  the  two  Foci. 


PROPOSITION  VIII. 

The  square  of  the  distance  of  the  Focus  from  the  cen- 
tre,   is  equal   to    the    dilFerence    of  the   squares    of  the 
semi-axes. 
Fig.  8.  That  is  CF2=CA2-Ca^ 

For  (111  Cor.  1.)  CA^    :  Ca^  y.CA^ -CF^-   :  FKS 
6  '»2  and  (19)  CA^    :  Ca^y.Ca^   :  FE2,(=|P2,) 

Therefore  CA2-CF2=Ca2, 

and  CF-^=CA2^Cfl2. 

Q.  £.  D. 


Fig.  9.      Cor.  1.— Hence FF2=AB2—a69=AB+fl6-AB-a6, 

or  F/is  a  mean  proportional,   between  the  sum,  and  dif- 
ference of  the  two  axes. 

Cor.  '2. — The  two  semi-axes,  and  the  Focal  distance 
fi'om  the  centre,  are  the  sides  of  a  right  angled  triangle, 
in  which  the  hypotenuse,  or  distance  of  the  Focus  from 
the  vertex  of  the  conjugate  axis,  is  equal  to  the  semi- 
transverse. 

_.     _      For     Fa=-Ca2=CF2, 
^^    •     and     CA2-Ca2=CF%.-.Fa=CA. 


OF  THE  ERLIPSE.  47 

Cor.  3. — The  square  of  the  Focal  distance  from  the 
centre,  is  equal  to  the  rectangle  of  the  semi-transverse, 
and  the  difference  of  the  semi-transverse  and  the  semi- 
parameter. 


That  is     FC^=CA  .  CA-FE, 

For     FC2=CA^"^C^and  aC^=(19)  AC  .  FE, 
Therefore  FC^  ^CA^  -AC  .  FE=AC  rAC"^^FE,  2.  i. 

and.'.     AC:FC::FC  :  AC-FE.  6.  i7. 

PROPOSITION  IX. 


The  sum  of  two  lines  drawn  from  the  Foci  to  meet  at 
any  point  in  the  curve,  is  equal  to  the  transverse  axis. 

That  is    /E4-/E=AB.  Fi-.  9. 

For,  draw  AG  parallel  and  equal  to  Ca  the  semi-conju- 
gate; and  join  CG  meeting  the  ordinate  DE  in  H  ;  also 
take  CI  a  fourth  proportional  to  CA,  CF,  CD  : 

Then(Il,CoR.1.)CA2  :  AG^ :  iCA^ -CD^  :  DE-%  6.22.4 

And,  (sim.  tri.)  CA2  :  AG^ ;  iCA^ -CD^  lAG^-DHS    ^- ^• 
consequently    DE^r^AG^  -DH2=Ca2  -DH^. 


Also       FD,  is  the  difference  between  CF.  and  CD, 
And       FD=^=CF=^--2CF  .CD-fCD2, 
FE-=FD2+DES 
Therefore   FE-^=CF2+C«=^ -2CF  .  CD-f  CD^^ -DH^ 


But  (VIII, )  CF2+Ca2=CA2; 

and  (Hyp.)   2CF  .  CD  =  2CA  .  CI  ; 

Therefore    FE2=CA=^-2CA  .  CI-f■CD2—DH^ 


2.  4. 


48  OF  THE  ELLIPSE. 

Again  (Hjp.)  CA^  :  CD^ :  :CF^(or  CA^^AG^^)  :  CI^  ; 
And         CA3  :CD«::CA2-AG2  :CD-"-DH^ 
CP=CD2-DH2, 
Consequentl}.F'E-=CA2-2CA  .  Cl-f  CI^. 


^  And   the   side   of  this   square   is  FE  =  CA  — CI=AI  ; 

..  4.  Cor.      jjj  ^^^  ^^^^  manner  it  is  found  that,/E=CA  +  Cl=BI  ; 
FE+/E==AI-fBI=:=AB. 

Q.  E.  D. 

Cor.  1  .—The  difference  between  the  semi-transverse, 
and  the  distance  from  the  Focus  to  any  point  in  the  curve, 
is  a  fourth  proportional  to  the  semi-transverse,  the  distance 
from  the  centre  to  the  Focus,  and  the  distance  from  the 
centre  to  the  ordinate  to  that  point. 

That  is  CA  :  CF::CD  :  CI.  (or  CA-FE), 

Cor.  2.— And /E-FE=2CI.  Oi*  the  difference  be- 
tween two  hues,  drawn  from  the  Focus  to  any  point  in  the 
curve  is  equal  to  twice  the  fourth  proportional,  to  CA, 
CF,  and  CD.     Therefore  CA  :  CF:  :2CD  :/E-FE. 


!oR.  3.— Hence  CA  .  /E  -  FE=t2CD  .  CF 

=(2CF  or)/F  .  CD. 


Cor.  4 Also/E.FE=CA-fCl.CA-Ci=CA2-CP. 

Cor.  5. — From  this  proposition  is  derived  the  com- 
mon method  of  describing  this  curve  mechanically,  by 
points,  or  with  a  thread  thus  : 


Fig.  Id.  In  the  Transverse  take  the  Foci  F,/and  any  point  L 
Then  w^ith  the  radii  AT,  BI  and  centres  F,  j\  describe 
arcs  intersecting  in  r„  which  will  be  a  point  in  the  curve. 
In  like  manner,  assuming  other  points  I,  as  many  other 
points  will  be  fouiid  in  the  curve.     Then  with  a  steady 


OF  T^E  ELLIPSE.  49 

iiand,  the  curve  line  may  be  drawn  through  all  the  points 
of  intersection    e. 

Or  take  a  thread  of  the  length  AB  of  the  Transverse 
axis,  and  tix  its  two  ends  in  the  Foci  F,  /  by  two  pms. 
Then  carry  a  pen  or  pencil  round  by  the  thread,  keeping 
it  always  stretched  and  its  point  will  trace  out  the  curve 
line. 


PROPOSITION  X. 


If  there  be  any  tangent,  and  two  lines  drawn  from  (he 
Foci  to  the  point  of  contact  ;  these  two  lines  will  make 
equal  angles  with  the  tangent. 


For  draw  the  ordinate  DE,  andy*e'  ;  parallel  to  FE,       Fig.  i2i 
Then(lXCor.l)CA  :  CD::CF  :  CA-FE  ;  p,  °^^ 

and  (VI)      CA  :  CD:  :CT  :  CA,  3 

therefore      CT  :  CFmCA  :  CA-FE, 
and  TF  :  T/::FE  :  2CA-FE,(or/E)-(IX);    5  e' 

But  (sim.  tri.)    TF  :  TF:  :FE  :/e', 

therefore     /E=/e' and  the  angle  f'  =  theangleyEe' ; 
but  because     FE  is  parallel  tofe',  the  angle  «'  =  the  an- 
gle FET  ; 
therefore     The  angle  FET=the  angle /Ee'. 

Q.  E.  D. 


ScHOL. — As  opticians  find  that  the  ailgle  of  incidence 
is  equal  to  the  angle  of  reflection,  it  appears  from  this 
proposition,  that  rays  of  light  issuing  from  one  focus,  and 
meeting  the  curve  in  every  point,  will  be  reflected  into 
lines  drawn  from  those  points  to  the  other  focus,  so  the 
ray/E  is  reflected  into  FE,  and  this  is  the  reason  why 
the  points  F,/,  are  called  ihe  foci  or  burning  points. 

Appollonius  called  the  Foci,  of  the  Ellipse  and  Hyper- 
bola Puncta  ex  comparatione  facta,  but  he  does  not  mention 
the  Focus  of  the  Parobala.  We  may  hence  conclude,  per- 
haps, as  well  as  from  the  term  Latus  Rectum,  (see  Prop. 
fK  Schol.)  that  the  latter  was  first  fixed  on,  and  that  the 


50  OF  THE  ELLIPSE. 

Foci,  were  discovered,  or  determined,  from  the  relation  of 
the  urdin  .tes  to  the  Latus   Rectum. 

N«^wton  and  others  called  the  Foci  Umbilici,  (Apollo- 
tiii  Conica — HamiUon's  Conic  Sections  102.  Newton's 
Princjpia  passim.) 

Cor.   1. — Hence  the  Normal^  EO,   or  the  line  drawn 
Fig.  9.  perpendicular   to  the  tangent,  from  the  point  of  contact, 
bisects  the  angle  made   by  the  two  lines,  drawn  from  the 
Foci. 

For  since  the  J^ormal,  makes  equal  angles  with  the  tan- 
ax  3  gent, 

OEP=OEF,  or  FE/is  bisected  by  EO. 

6.  3.        Cor.  2.-— Consequently  FE  :/E:  :F0  :/0  ; 
^■^'  and  FE+/E  :/E  -  FE:  :F0+/0  :/0- FO  ; 

That  is  (IX)  2CA  :  2C1:  :F/  :  2C0, 
therefore      CA   :    C1::CF:C0. 

CoR.  3.-Therefore  CA^  :  CA  .  CI:  :CF-  :  CF  .  CO, 
and      CA^  :  CF^ :  :(CA  .  CI=)  CF  .  CD  :  CF  .  CO, 
.-.      CA^  :  CF2::CD  :  CO. 

CoR.  4.-Therefore  CA^  -CF^  :  CF^ :  :CD-CO  :  CO. 
thatis(Vlll)         aC-     :CF2::D0:C0. 

CoR.  5.— CA«  :  CA2-CF-::CD3  :  CD-CO. 
that  is      CA2  :  Ca2::CD  :  DO. 

6.20.        CoR.  6. -Hence  (19)  AB:P::CD:DO, 
^''''^'  and  AC  :iP::CD:DO, 

AC.  DO=iP.  CD. 
And  if,  as  in   the    Parabola,  AC  and  CD,  be  supposed 
equal,  then  DO,  the  subnormal^  equals  half  the  Parame- 
ter. 

CoR.  7.-2CA  :  2CI::/F:  2C0::CF  :  CO, 
2CA  :  2CA-2CI:  :CF  :  CF-CO. 
thatis(IX)  2CA:  2FE  ::CF:OF; 

or  CA:FE  ::CF:OF. 


OF  THK  ELLIPSE.  51 


PROPOSITION  XI. 


If  a  line  be  drawn  from  either  Focus,  perpendicular  to 
a  tangent  to  any  point  of  the  curve,  the  distance  of  their 
intersections  from  the  centre  will  be  equal  to  the  Semi- 
transverse  axis. 

That  is  CP  and  Cp  each  =  CA  or  CB.  Fig.  12, 

For  throtigh,  the  point  of  contact  E  draw  FE,  and/E 
meeting  FP  produced  in  G,  then  the  angle  GEP  =  dngle 
FEP,  being  each  equal  to  the  angle/e/?,  and  the  angles  at 
P  being  right  and  the  side  PET  being  common,  the  two 
triangles  GEP.  FEP,  are  equal  in  all  respects,  and  so 
GE  =  FE,  and  GP  =  FP.  Therefore,  since  FP  =  iFG, 
and  FC  =  |F/,  and  the  angle  at  F  common,  the  side  CP 
will  be  ^iFG  or  lAB  that  is  CP  =  CA  or  CB. 

And  in  the  same  manner  C  2?=CA  or  CB. 

Q.  E.  D, 

CoR.l. — A  circle  described  on  the  Transverse  axis,  will 
pass  through  the  points  of  intersection  P  and  p. 

CoR.  2. —  If  from  the  intersections  of  a  tangent  with  a 
circle,  circumscribing  an  Ellipse,  perpendiculars  be  drawn 
they  will  pass  through  the  Foci. 

Cor.  3. —The  distances  of  the  Foci  from  the  point  of 
contact,  are  to  each  other  as  the  perpendiculars. 

For  the  triangles  EPE,yyoE  are  similar  ; 
Therefore  FE   : /E:  :FP  : //?. 

Cor.  4.-If  PF  and  pC  be  produced,  they  will  meet    in 
the  circumference  of  the  circle  in  K,  of  which  /?K  will  be 
the  semi -diameter,  for/PK  is  a  right  angle, 
also  FKC,  C;?/*are  similar  and  equal  triangles  ; 
Therefore  PF  .  pf=?V  .  FK  =  AF   .  FB=Co*.  (Vlf. 
Cor.  5.) 


5^  OF  THE  ELLIPSE. 


PROPOSITION  XII. 

The  distance  of  the  Focus  from  any  point  in  the  curve, 
is  equal  to  the  ordinate  to  that  point,  continued    until   it 
meets  the  Focal  tangent, 
Fig.  n.  that  is  Fa==C^  FA=AL,  FB=Bz,  and  generally  FM=» 
DG, 
for(19)  FE  =  iP, 

and  (VII  Cor.  5)  FE  .  C/=iPAC=FE  .  AC, 
therefore  C^=AC=Fa.  (VIII) 

Again  (VI)  CF  :  AC: :  AC  :  CT, 
and        AC  :  AF::CT  :  AT, 
Sim.  tri.       AT  :  TC::AL  :  C<,  =  CA  ; 

AF  :  AC::AL  :  AC, 
therefore    AF=AL. 

That  is,  the  distance  from  the  vertex  to  the  Focus,  is 
equal  to  the  tangent  lo  the  vertex,  intercepted  by  the 
Focal  tangent. 

(VI  Cor.  7,)    AF  :  FB:  :AT  :  TB, 
.-.Sim  tri.       AT  :  TB: :  AL  (=AF)  :  BZ, 
therefore      AF  :  FB:  :AF  :  BZ, 
BZ=.BF. 


(VI.)  AC^  =CF .  CT=CF :  CF+FT=CF2  -|-CF  .  FT, 

.♦.  AC2-CF~^(=aC^)=CF.FT, 
Sim.  tri.  TF  :  FE:  :TC  :  C/.  =  AC, 
but    TC  :  AC:  :TC  .  CF  :  AC  .  CF, 
-  .     TC.CF(=AC2)  :  AC  .CF::AC:CF, 
that  is  no  TF  :  FE : :  (AC  :  CF) : :  TA  :  AL : :  TD  :DG  ; 
but  (IX)  CT  :  AC::CD  :FM-CA, 
and       CT+CD  :  AC  +  FM-CA:  :CT  :  AC, 
that  is    TD:  FM::CT:  AC; 

.-,      TD  :  DG:  :TD  :  FM.-.DG=FM. 

Q.  E.  Z). 

CoR.l — If  through  T,  a  line    be  drawn   perpendicular 
to  TA,  it  is  called  the  directrix. 


OF  THE  ELLIPSE.  53 

The  distance  of  any  point  in  the  curve,  from  the  direc- 
trix as  My  IS  equal  to  TD  which  has  a  constant  ratio  to 
DG  oiF-M,  the  distance  of  the  point  from  the  Focus.  In 
the  Parabola  it  is  a  ratio  of  equality. 


PROPOSITION  XIII. 

If  tangents  and  ordinates  be  drawn  from  the  extremities 
of  any  two  conjugate  diameters,  meeting  the  Transverse 
axis  produced,  the  distances  of  these  intersections  from 
the  center  are  reciprocally  proportional* 

That  is    CD  :Cd:\Ct:CT,  Pi?- J 4. 

For(V[)    CD  :CA::CA  :CT, 
and  ,,     Cd  t    CA;  :CA  J  C/, 
CD  :Cd::Ct  :  CT. 

Q.  E.  D. 

Cor.  1. — Hence  the  distance  from  the  center,  to  the 
ordinate,  at  the  extremityfof  one  diameter,  is  a  mean  pro- 
portional between  the  distances  of  the  intersections  of 
the  tangent  and  ordinate  drawn  from  the  extremity  of  the 
other. 


For  since      CD  :  Cdy.Ct  :  CT, 

and  bysim.  tri.  Cd  :  DT:  :C^  :  CT, 

CD  :Cd::Cd:  DT. 

In  like  manner  Cd  :  CD:  :CD  :  dt. 

Cor.  2  — The  ordinates,  also,  are  reciprocally  as  their 
distances  from  the  center. 

Forsim.  tri.     DE  :  de.'.Cd:  CD. 

CoR.  3. — Therefore  the  ordinates  are,  as  the  distan- 
ces of  the  intersections  of  their  tangents  from  the  center, 
or  (Cor.  1,)   DE  ideWCT  :  Ct . 

8 


54  OF  THE  ELLIPSE. 

Cor.  4. -Also  by  (Cor.  2,)  The  rectangles  DE  .  CD 
therefore,  the  triangle  CDE=-Crff. 


PROPOSITION  XIV. 

If  ordinates  to  the  Transverse  axis,  be  drawn  from  the 
extremities  of  any  two  conjugate  diameters,  the  sum  of 
their  squares  is  equal  to  the  square  of  the  Semi-Conjugate 
axis,  and  the  sum  of  the  squares  of  two  ordinates,  drawn 
from  the  same  extremities  to  the  Conjugate  axis,  is  equal 
to  the  square  of  the  Semi-Transverse. 

Fig.  14.       That  is,         I>e^-\-de^=CaK 
And  D'e'-{-d'e^=CA\ 

For  (VI)      CD  :  CA::CA:CT, 

CD  :CA::AD  :  AT  ; 
And  CD:DB::AD:DT, 

.  .  (XIII.  2)  CD  .  DT  (=Crf2)=AD .  DB=CA2  ~CD^ 

CJ2=CA=^-CD2; 
And  CD2  4-C(/2=CA2; 

That  is  D'E^-+d'e'=CAK 

In  the  same  manner  DE=-|-c?'«2=Ca2. 

q.  E,  D. 

CoR.  1.— Hence    AD  .DB=C(;%  and  AJ.  c/B=CD^ 

Cor.  2.— Hence     CD^  +0^^  =CA^ ; 
and         CD'2  4-C(i'^=Ca2; 

Cor.  3.— Since     CX'  :Ca' '..{AD  .DB=)  Cd^:DE\ 
.'.       CA:Ca::Crf:DE. 
and       CA:Ca::CD:de. 


OF  THE  ELLIPSE.  ^$ 


PROPOSITION  XV. 


If  from  the  extremity  of  any  diameter  a  perpendicular 
be  drawn  to  its  conjugate,  the  rectangle  of  this  perpendic- 
ular, and  the  part  of  it  intercepted  by  the  Transverse  Axis, 
is  equal  to  the  square  of  the  Semi-Conjugate  axis. 


Fig:.  14. 


That  is        EP.EI=Ca^ 

Draw  C^  parallel   to  EP,  the  triangle  EDI,  and  Cyt  are 

similar  : 
Therefore  Ct  :  Cyr.El :  ED, 

That  is  Cr  :  EP::EI:C(/', 

Therefore  EP  .  EI=C/  .  Cd=^Ca'. 

Q.  E.  Z>. 


PROPOSITION  XVI. 

All  parallelograms  circumscribed  about  an  Ellipse 
whose  sides  are  parallel  to  two  conjugate  diameters,  are 
equal  to  each  other,  being  each  equal  to  the  rectangle  of 
the  two  axes. 

That  is        KQNS=RYZX=AB  .  ab.  Fig.  13. 

For  (II  h  XIV)  CA^ :  CaA  ! (AD  .  DB  or)  C^/^  :  DE^; 
Therefore  C A  iCay.Cd:  DE, 

In  like  manner    CA  :  Ca !  :CD  :  de  ; 
Or  CA:CD::Ca:de. 

But  (VI)  CA:CD::CT:CA, 

Ca  :  del  :CT  :  Ca, 
Also,  siui.  tri.  Ce  :  de:  ;CT  :  Cy, 
Therefore  Ca  iCe'.'.Cy  :  CA  ; 

And  Ca  .  CA=C«  .  Cy  or  G^  .  EP. 

4Ca  .  CA,  or  AB  .  ab=iCe  .  EP,  or  QKSN. 

Q.  E.  D, 


55  OF  THE  ELLIPSE. 


PROPOsrrioN  xvii. 

The  sum  of  the  squares  of  every  pair  of  conjugate  di- 
am  ers,  equal  to  the  same  constant  quantity,  viz.  the 
sum  of  th-  squares  ol  the  two    xes. 

That  is,  AB  -{-ab^  =EG'  -{-egK 

Fig.  14.      For  CE2+Ce2=-CD2+CJ2-fDE2-fDe2; 

But   XIV,      D  +Cri^=AC3;andDE^-fr/e^=Cfl2; 
therefore       CE   -|-Cc  ^AC^+Cr-', 
and  EG''+eg^=AB^-]-ab\ 

Q,  £.  A 


PROPOSITION  XVIII. 

If  the  extremities  of  the  transverse  and  conjugate  axes 
be  connected,  the  diameters  which  are  parallel  to  the  con- 
necting line,  are  conjugate  and  equal  to  each  oiher. 

Fig.  16.      That  is,         EL=GK,  and  they  are  conjugate  to  each 

other. 

For  Aa=«B=B  =6A.*.AaB6,  is  a  parallelogram, 

also  the  angles  GCB=a\B=hAB  =  LCB, 
and  since  AB=2CB.  .Ba=2BR, 

aB  is  a  double  ordinate  to  GK, 
and  EL  parallel  to  it,  is  conjugate  to  GK. 

Also  CG=CL  or  KG=EL. 

Q,  E.  D. 

Corollary. — These  are  the  only  two  congugate  di- 
ameters in  the  Ellipse,  which  are  equal. 

This    Corollary  and   the  last  step  in    the   demonstration 
may  easily  be  demonstrated   by  reductio   ad   absurdum. — 
See     Emerson's    Conic    Sections — Ellipse — Prop.    3.5, 
36.  37. 


OF  THE  ELLIPSE. 


PROPOSITION  XIX. 


57 


If  from  any  point  in  the  Ellipse,  a  line  be  drawn  to  the 
conjugate  axis,  equal  to  the  semi-transvere  axis,  the  part 
of  the  line  intercepted  between  that  point,  and  the  trans- 
verse axis,  is  equal  to  the  semi-conjugate. 

That  is,  if    MG=AC,"then  MI=aC.  Fi^.  s. 

For,  sim.  tri.     MG  :  MR:  :MI  :  ID, 

and  MG2  :MR   ::MI-^  :  IDS 

that  is  AC2  :  ( :D»::MP  :  ID^, 

and  AC^  :  AC^-CD-.iMP  ;  MI'-ID^, 

AC  :  AD.DB::  MI^iMD, 
AC--  :MI^::AD.DB:MD2  ; 

but  (II)  AC-  :  «C2 : :  AD  .  DB  :  MD% 

MV=aC^  and  MI=aC. 

Q.  E.  a 

CoR.— Conversely— If  ]MI=aC  then  MG  =  AC:. 


PROPOSITOIN  XX. 

If  a  tangent  and  ordinate  be  drawn  from  the  extremi- 
ties of  the  transverse  axis,  and  of  any  other  diameter,  and 
be  produced  to  meet  that  diameter  and  transverse  axis,  the 
triangles  formed  respectively  by  these  lines,  are  equal. 

That  is.    CAE=CMT  and  AGT=MGE,  Fig.  17. 

also       CHN=CDiM  and  AHE=MDT. 

1.  For  CDM  and  CAE,  also  CHA  and  GMT  are 

similar. 
.;.  CH  :  CM::CA  :  CT, 

::CD  :  CA, 

::CM  :  CE;  -^ 

therefore  DH,  AM  and  TE  are  parallel; 

AME  =  AMT,  and  CAE  =  CMT. 

2.  Also  taking  CMGA  from  each,  AGT=MGE.  6.  15. 

3.  *    adding  MDA,  to  each,  xMDT  =  AHE. 

4.  therefore      CDM=CHA. 

Q.  E.  D. 


d» 


OF  THE  ELLIPSE. 

CoR.l.— AME+AMD=MDAE=MDT  ; 
AMH4-AMT= AHMT=AEH ; 
MDAE=AHMT. 

Cor.  2.         QIR=AEFI. 

Forsim.tri.CA;AE::CD:DM::CD-fCA:DM-fAE; 
and  CA  :  AE::CI  :  IF::Cl-|-CA  :  IF+AE; 


Again       AI .  IB=AI .  IC+CA  ; 


and  AD.DB=AD.DC-fCA;  

therefore  AI. IB  :AD.DB::AI.IC4-CA:  AD.  DC+CA, 


:AI.AE+1F:AD.AE+DM, 
.2AEFI:  2AEMD, 
:   AEFl  :  AEMD. 
Butsirn.tri.  QIR  :  MDTliQi^   :  MDS 

::AI  .  IB  :  AD  .  DB, 
::AEFI  :  AEMD; 
But  (Cor.  1.)     MDT=AEMD  .  .  QIR=AEFI. 

Cor.  3.         QLF=LMTR. 

For  take  CLR  from  CAE  and  from  CMT, 
then     LRTM=LRAE=LRIF+AEF1=LRIF-|-RQI 
=L(iF. 


PROPOSITION  XXI. 

The  square  of  any  diameter, 
Is  to  the  square  of  its  conjugate, 
As  the  rectangle  of  its  abscisses, 
Is  to  the  square  of  their  ordinate. 

Fig.  17.      That  is        NM^  :  VK^iiNL  .  LM  :  LQ^ 

For  (XX.  Cor.  3.)         LQF=LRTM ; 

and  as  LRQ  moves  parallel  to  itself,  towards  CK, 

LRTM  approaches  to  equality  with  CTM ; 
They  may  therefore  be  considered  as  ultimately  equal, 
and    CKO=CMT. 


OF  THE  ELLIPSE.  59 

But  Sim.  tri.  CMT  :  CRLr.CINP  :  CL^  6.  20. 

and         CMT  :  CM  r-CKL:  :C]VP  :  CM^— CL*, 
that  is       CMT  :  LRTMxCM^   :  NL .  LM  ; 
Butsim.  tri.  CK-'   :  LQ^iiCKO  :  LQF, 

::CMT:  LRTM, 
::CM'  :  NL.LM, 
therefore     CK=   CM^  :  iLQ^   :  NL  .  LM  ; 
and         NM^   :  VK^::NL.LM  :  LQ^ 

Q.  E.  D, 

ScHOL. — The  preceding  proposition  shows,  that  any  di- 
ameter of  an  Ellipse,  and  its  rectangles,  have  the  same  re- 
lation to  its  conjugate,  and  ordinates,  as  have  been  demon- 
strated in  the  case  of  the  Axes.  All  the  properties  therefore, 
which  have  been  deduced  from  this  relation,  in  the  former 
case  in  reference  to  lines  intersecting  each  other,  parallel  to 
the  axes,  and  to  tangents  intersecting  these  lines  and  the 
curve,  may  be  applied  to  any  conjugate  diameters.  The 
first  six  propositions,  therefore,  of  this  chapter,  with  their 
corollaries,  are  particular  truths  which  may  safely  be  gen- 
eralized by  applying  them  to  any  two  conjugate  diameters, 
as  well  as  to  the  two  axes.  The  general  propositions, 
however,  will  be  stated,  and  the  corresponding  figures 
have  been  drawn  and  lettered  in  a  similar  manner,  so  that 
the  demonstrations  already  given,  for  the  axes,  are  appli- 
fjable,  with  slight  variations,  to  those  propositions  also, 
which  will  be  arranged  like  the  corresponding  propositions, 
concerning  the  axes. 

CoR.  L — The  squares  of  the  ordinates  o{  any  diameter 
are  to  each  other,  as  the  rectangles  of  their  abscisses. 

That  is  CK^  :  LQ-^ :  :NC  .  CM  :  NL  .  LM.  (See  I  Prop.)  Fig.  17. 

Cor.  2. — The  corresponding  ordinates  on  each  side  of 
any  diameter  are  equal. 

CoR.  3. — Ordinates  at  equal  distances  from  the  center, 
or  from  the  vertices  of  any  diameter  are  equal ;  and,  con- 
versely, equal  ordinates  are  at  equal  distances  irom  the 
center  and  from  the  vertices. 


60  OF  THE  ELLIPSE. 

Cor.  4. — The  whole  Ellipse  is  divided  by  every  diam- 
eter into  two  equal  parts,  which  may  be  placed  so  as  to 
coincide  in  every  respect. 

Cor.  3. — Therefore  the  parts  of  the  tangents,  at  the  ex- 
tremities of  the  diameter,  would  coincide,  and  the  two  tan- 
gents are  parallel,  as  demonstrated,  Prop.  1  and  III. 

Cor.  6. — Any  diameter  is  to  its  parameter,  as  th  e  rec- 
tangle of  its  abscisses,  is  to  the  square  of  their  ordinates. 

Cor.  7. — The  rectangle  of  the  absciss  of  any  diameter 
and  its  parameter,  exceeds  (he  square  of  the  ordinate,  by 
a  rectangle  similar,  and  similarly  situated  to  the  rectangle 
of  the  diameter  and  its  parameter. 

That  is     DB  .  BG>DE-^  by  FIHG. 

CoR.  8. — If  a  circle  be  described  on  any  diameter, 
The  ordinate  in  the  circle. 
Is  to  the  corresponding  ordinate  in  the  Ellipse. 
As  the  given  diameter, 
To  its  conjugate. 
Fig.  18.     That  is  CA  :  Ca:  :DG  :  DE. 


PROPOSITION  XXll. 

The  square  of  any  diameter,  is  to  the  square  of  its  con- 
jugate, as  tile  rectangle  of  the  segments  of  a  line  in  the 
Ellipse,  parallel  to  the  former,  is  o  the  corresponding  rec- 
tangle of  the  segments  of  a  line  parallel  to  the  latter. 
19.  That  is,  CA=^  :  Ca2  or  AB^  :  ah^  :\eh  ,  hg-\-Eh  .  ^G. 
SeeProp.  IV,  &ic. 

Cor.  1. — Therefore  by  equality  of  ratios  : 
The  corresponding    rectangles  of  the  segments  of  lines 
intersecting  each  other,    parallel  to  any   diameter   and   its 
conjugate,  have  to  each  other,  always  the   same  ratio,  viz. 
that  of  the  squares  of  their  diameters. 

That  is,  eh.hgiek,  kg:  :Eh  .  ^G  :  ik  .  kl. 


OF  THE  ELLIPSE.  gj 

ScHOL. — If  the  Ellipse  become  a  circle,  then  the  same 
proposition  holds  true,  as  is  demonstrated  by  Euclid,  but 
the  ratio  is  then,  by  alternating  the  terms,  a  ratio  of  equal- 
ity. Ill  Book,  35. 

l(  hgf  and  kg,  are  considered  equal,  as  in  the  Parabola, 
then  the  proposition  \s  eh  :  ekwE^h  .  AG  :  ik  .  kl. 

As  demonstrated  of  the  Par.  Prop.  II. 


PROPOSITION  XXIII. 

If  lines  in  the  Ellipse  parallel  to  two  conjugate  axes,  in- 
tersect each  other  without  the  Ellipse,  the  rectangles  of 
corresponding  segments,  are  to  each  other  as  the  squares 
of  the  diameters  to  which  they  are  parallel. 

That  is,  ;  B^  :  ab^ :  :eH  .HglEH  .  HG. 

See  Prop.V,  &;c. 

CoR.  I. — Hence  the  rectangles  of  the  segments  of  all 
lines,  parallel  to  two  conjugate  diameters,  and  intersecting 
each  other  without  the  Ellipse,  have  to  each  other  always 
the  same  ratio,  being  the  ratio  of  the  squares  of  those  two 
diameters. 

That  is,  EH  .  HG  :  eH  .  H^:  :K1  .  IL  :  EI  .  I/. 

Cor.  2.— Also  EH  .  HG  :  ES  .  SG:  :eH  .  H^  :  Sa^ 
€oR.  3.— Also  ek.kniEk  .  kl:  :EH  .  HG  :  cH  .  Bg. 


PROPOSITION  XXIV. 

The  squares  of  any  two  diameters,  are  to  each  other,  as 
the  rectangles  of  the  segments  of  one  of  those  diameters, 
or  of  lines  parallel  to  it,  are  to  the  corresponding  rectangles 
of  the  segments  of  lines  parallel  to  the  other. 

That  is  AB3  :  EG^ : :  AD  .  DB  :  KD  .  DL.  Fig. 

9 


62  OP  THE  ELLIPSE. 

There  is  of  course  no  corresponding  proposition,  in  the 
caseof  the  two  axes,  as  they  have  a  definite  position  in 
regard  to  each  other,  but  this  more  general  proposition 
may  be  deduced  from  the  preceding  propositions  by  the 
ordinary  changes  of  geometrical  ratios. 

CoR.—ed  .dg:Kd,  dL::CB^  :  CG' : : AB^ : EG», 

and        ed  .  dg  :  eh  .  hg:  '.Kd  .  dLi  :  Eh  .  AG. 

That  is,  the  rectangles  of  the  corresponding  segments  of 
all  parallel  lines,  intersecting  each  other  in  the  EllipsCy 
have  to  each  other   the  same  ratio. 


PROPOSITION  XXV. 

If  any  two  lines,  cutting  the  Ellipse,  intersect  each  oth- 
er, without  it,  the  rectangles  of  their  segments,  are  to  each 
other,  as  the  squares  of  the  diameters,  or  semi-diameters  to 
which  thevare  parallel. 
Fig.  23.      That  is  EH  .  HG  :  eH  .  H^:  :CG'-^  :  C^'^ 

This  is  deduced  in  a  manner  similar  to  the  last, 

CoR.  1. — The  squares  of  tangents  intersecting  each  oth- 
er, are  to  each  other,  as  the  squares  of  the  diameters,    or 
semi-diameters,  to  which  they  are  parallel ;  or  the  tangents, 
are  as  the  parallel  diameters. 
Fig.  22.      That  is      HxM-^  :  HN2  yCm^  ♦  Cn^  ; 
or  HM  :  HN::Cm  :  Cw. 

rig.  22       Cor.  2.— IQ  .  QS  :  ED  .  DL:  :T)N'  :  QM% 

andlQ.QS  :ED  .  DL:  :IP  .  PSiEP.PL:  iCG^  :CK-. 

Cor.  .3.— HN  :  HM : :  TF  :  TM  : :  CG  :  CK, 
and  LR.RE-.RF^I'.ED.  DLiDN^iCK^  :  CG^. 

ScHOL. — In  the  two  preceding  propositions,  with  their 
Corollaries,  the  principle  involved  in  the  preceding  gene- 
ral propositions,  (and  of  course  in  the  particular  prop osi- 
lions  relative  ro  tlr?  axis,)  is  seen  in  its  most  general  form. 
It  may  be  expressed  as  follows.     If  parallel  lines  in  the 


OF  THE  ELLIPSE.  G3 

Ellipse  intersect  another  line,  or  intersect  other  parallel 
lines,  either  within  or  without  the  Ellipse,  the  rectangles  of 
their  corresponding  segments,  ore  always  in  the  same  ratio, 
and  if  any  of  the  lines  touch  the  Ellipse,  the  squares  of 
the  intercepted  parts  of  those  lines  may  be  considered  as 
rectangles,  and  are  as  the  squares  of  diameters  parallel  to 
them. 

It  will  be  seen  by  trial,  (reccollecling  that  the  square  of 
a  semi-diameter,  or  of  an  ordinate  is  equal  to  the  rectangle 
of  the  equal  parts  of  the  whole  diameter,  or  double  ordi- 
nate) that  this  general  proposition  as  now  stated,  is  appli- 
cable to  all  the  cases  alluded  to,  and  includes  propositions 
I,  II,  III,  IV,  V,  XXI,  XXII,  XXIII,  XXIV,  XXV, 
From  it  also  may  be  very  easily  deduced,  all  the  general 
propositions  which  are  most  useful,  concerning  diameters, 
tangents,  he. 

PROPOSITION  XXVI. 

If  a  tangent  and  ordinate  be  drawn  to  any  point  in  the 
curve,  meeting  any  diameter,  the  half  of  that  diameter  is 
a  mean  proportional  between  the  distances  of  the  two  in- 
tersections from  the  centre. 

That  is  CD.CT=CA2, 
or         CD  :  CA  :  CT,   are  continued  proportionals. 

{This  may  be  demonstrated  in  the  same  manner  as  in 
Prop.  VI  but  it  is  immediately  deduced  from  the  Gene- 
ral Ppoposition  just  stated,  as  it  is  illustrated  in  Prop. 
XXV,  Cor.  3.) 

For  HB:AK::EH  :EK::BD  :DA::BT:  AT;         Fig.  24. 
andBD-DA  (or  2  CD)  :  BD:  :BT  -  AT  (or  BA) :  BT, 
CD  :BD::BC  ;BT, 
and         CD  :BD-CD::BC:BT-BC; 
That  is    CD  :  AC::  AC  :  CT. 

q.  E.  D. 

CoR.   1.— The  two  angents,  ET,  ET  drawn  at  the  two  ^'^S-  25. 
extremities  of  the  same  double  ordinate,  intersect  the  diam- 
eter in  the  same  point. 

And,  conversely,  if  a  diameter  pass  through  the  point  of 
intersection,  it  will  bisect  the  line  which  connects  the  points 
of  contact. 


64  OF  THE  ELLIPSE. 

Fig.  24.      Cor.  2.— HK  and  FG  are  bisected.     Hence  IF=MG. 

Cor.  3. — DT  is  harmonically  divided,  or 
DT  :  AT::BD  :  AD. 

Cor.  4.— TA  :  TD:  :TC  :  TB. 
also  TK  :  TE::T^  :  TH. 

and  AK:  DE::C/-fBH. 

ScHOL. — As  the  Transverse  Axis,  passes  through  the 
Foci,  it  has  particular  relations  to  those  points,  which  no 
other  diameter  can  have.  In  the  relations  which  other  di- 
ameters have  to  lines  drawn  through  the  Foci,  there  are 
however  some  analogies  to  those  of  the  axis,  as  will  be 
seen  in  the  Notes. 


OF  THE  HYPERBOLA. 


PROPOSITION  I. 

The  squares  of  any  two  ordinates  to  the  axis,  are  to 
each  other,  as  the  rectangles  of  their  abscisses. 

Let  MVN  be  a  triangular  section  through  the  axis  of  the  ^^n-  *• 
cone  (3),  AGIH,  another  section  perpendicular  to  the  for- 
mer, forming  an  Hyperbola;  AH,  the  mutual  intersection  of 
the  two,  will  be  the  transverse  axis  of  the  Hyperbola  (11); 
let  MIN,  KGL  be  circular  sections  parallel  to  the  base  (4) ; 
FG,  HI,  the  mutual  intersections  of  these  planes  with  that 
of  the  Hyperbola,  will  be  ordinates,  both  in  the  Hyperbola, 
and  in  the  circles,  being  perpendicular,  both  to  MN  and 
KL,  and  to  AB. 

Then    FG^  :  HP  : :  AF  .  FB  :  AH  .  HB. 

For  Sim.  tri.  AF  :  AH : :  FL  :  HN, 

And  FB:HB::KF  :MH; 

Therect. .:.  AF  .  FB  :  AH  .  HB::KF.  FL  :  MH  .  HN,   6.  C. 

But  KF  .  FL=FG^  and  MH  .  HN=HIS  _  .  _ 

Therefore  AF  .  FB  :  AH  .  HB:  :FG=»  :  HP. 

Q.  E.  D.  5.  7. 

ScHOL. — If  then  a  straight  line,  placed  perpendicularly 
\ipon  another  finite  straight  line  produced,  were  to  move 
parallel  to  itself,  and  to  vary  in  its  length  continually, 
60  that  its  squares  should  always  have  the  same  ratio 
to  each  other,  as  the  rectangles  of  the  parts  between  it 
and  the  extremities  of  the  other  line,  the  figure  formed  by 
the  moving  line,  would  be  the  portion  of  an  Hyperbola 
between  the  axis  and  carve. 


6S  OP  THE  HYPERBOLA. 

Cor.  1. — As  the  proposition  and  demonstration,  are 
equally  applicable  to  ordinates  on  opposite  sides  of  the 
axis,  having  the  same  abscisses,  those  ordinates  are  equal 
or  GS,  is  double  of  FG.  Hence  it  is  called  a  double  or- 
dinate. 

Cor.  2. — The  squares  of  the  ordinates  are  the  same 
as  the  rectangles  of  the  equal  segments  of  the  double  or- 
dinates, therefore,  Ihe  proposition  may  be  thus  expressed. 
The  rectangles  of  the  segments  of  the  double  ordinates^  are 
to  each  other  as  the  rectangles  of  the  abscisses,  of  the  tranS' 
verse  produced. 

Cor.  3. — Ordinates  at  equal  distances  from  either  ver- 
tex, are  equal  ;  since  the  abscisses  and  consequently  the 
rectangles  of  the  abscisses  are  equal.  And  conversely,  if 
the  ordinates  are  equal,  their  distances  from  the  vertices 
are  equal, 

CoR.  4. — Hence  the  Foci  are  equally  distant  from  eith- 
er vertex,  for  their  ordinates  are  equal,  being  each  equal 
to  the  semi-parameter,  (20),  their  distances  from  the  cen- 
tre are  therefore  equal. 

CoR.  5. — Hence  also  every  diameter  is  bisected  in  the 

centre.     For  since  at  equal  distances  from  the  centre,  the 

T\%.  13.  ordinates  are  equal,  the  third  sides  and  the  remaining  an- 

1.  14.  ^^^  °^  *^^  T\g\\\.  angled  triangjes,  CDE,  Cd'G  are  equal, 

therefore  ECG,  is  a  right  line  bisected  in  the  centre. 

And  conversely,  the  ordinates  from  the  ends  of  any  diam- 
eter upon  the  transverse  axis  are  equally  distant  from  the 
centre. 

CoR.  6. — Hence  also,  the  whole  Hyperbola,  is  by  the 
transverse  axis,  divided  into  two  equal  parts,  which  if  pla- 
ced one  upon  the  other  will  coincide  in  every  respect. 
For  if  they  should  not  coincide  in  any  part,  the  or- 
dinate in  one,  at  that  point,  would  be  unequal  to  the  cor- 
responding ordinate  in  the  other. 

CoR.  7. — For  similar  reasons,  the  two  parts  of  the  tan- 
gent at  the  vertex,  would  coincide  ;  that  is,  the  tangent  at 
the  vertex  makes  equal  angles  with  the  axis,  on  each  side 
of  it.     It  is  therefore  perpendicular  to  the  axis. 


OF  THE  HYPERBOLA.  (57 

Cor.  8. — The  two  opposite  sections  of  a  Hyperbola, 
are  equal  and  similar,  and  if  placed  upon  each  other  would 
coincide  in  every  respect ;  for.  ag  the  ordinates  at  equal  dis- 
tances from  either  vertex  are  equal,  the  points  in  the 
curve  which  limit  these  ordinates  would  coincide. 

Cor.  9.  —  The  two  diameters,  drawn  from  the  extrem- 
ities of  any  double  ordinate  to  the  axis,  are  equal. 

Cor.   1 0. — If  a  circular  section  as  POC,  pass  through  the  ^'•0-  i- 
centre  of  the  Hyperbola,  the  semi-conjugate  Ca,  is  a  mean 
proportional  between  OC  and  CP. 

For  the  whole  conjugate  axis,  is  a  mean  proportional 
between  BR  and  AQ  ,  the  diameters  of  the  circular  sec- 
tions, passing  through  the  two  vertices  of  the  Hyperbola. 

(13.)  or     BR  :  aby.ab:  AQ, 

but       AB=2AC,.-.BR=2CP,  and  QA=20C, 
therefore  OC  :  Ca'.'.Ca  :  CP.  5,  15. 

SciiOL. — It  has  been  already  remarked,  that  if  the  cut- 
ting plane  be  supposed  to  revolve  about  A,  the  axis  AB  will 
become  more  extended,  the  abscisses  FB  and  HB,  will  ap- 
proach to  a  ratio  of  equality,  and  the  section  itself  will  ap- 
proach indefinitely  near  to  a  Parabola  ;  and  when  it  be- 
comes parallel  to  the  side  of  the  cone,  it  will  be  a  Parabo- 
la (5).  If  then  the  indelinite  abscisses  are  considered  Fig.  2. 
equal,  the  precedintj  propositiofi  will  become, 

AF  :  AH::FG*::HP  e.  i. 

the  same  as  the  first  proposition  of  the  Parabola. 

Again  if  the  cutting  plane,  move  parallel  to  itself,  on  the 
base  MN,  towards  M,  A  and  B  will  approach  continually 
towards  V,  FB  and  HB  will  also  approach  to  equality  with 
FA  and  HA.  When  A  and  B  coincide  in  V,  the  section 
will  become  a  triangle  (3)  and  FB,  HB  will  be  equal  to 
FA,  HA,  and  the  rectangles  AF  .  FB,  AH  .  HB,  will  be 
equal  respectively  to  AF^,  and  AH^,  then  the  proposition 
becomes 

AF2  :  AH^::FG2  :  Hi% 

AF    :  AH    :  :FG    :  HI.  (See  Euclid,  6,  1.) 


68  OF  THE  HYPERBOLA. 


PROPOSITION  II. 


As  the  square  of  the  Transverse  Axis, 
Is  to  the  square  of  the  Conjugate  Axis, 
So  is  the  rectangle  of  the  abscisses  of  the  Transverse, 
To  the  square  of  their  ordinate. 
Fig  1-      That  is  AB^  :  a6»  : :  AD  .  DB  :  DE^. 

For  sim.  tri.  BC  :  CO:  :BF  :  FK, 
and    „      „    AC:CP::AF:FL, 

AC  .  CB  :  CO  .  CP:  :AF  .  FB  :  KF  .  FL ; 
that  is       AC2  :  Ca^ : :  AF  .  FB  :  FG=»  ; 
so       AC^  :Ca2::AD.DB:  DES 

5.  15.  or      AB2  :  ah^ : :  AD  .  DB  :  DE^ 

q.  E.  D. 

2.  6.        Cor.  1.— Since  the  rectangle  AD  .  DB:=CD2  -CA% 

Therefore     AC^  :  aC^ :  :CD»  -CA^  :  DE^; 
or  AB^  :  ab^  ::CD»— CA^  :  DE^. 

CoR.  2.— (19)AB    :ah    lab    :  Parameter=P, 

cof'2        ^^^^        ^^   *  ^    ::AB2  :  a6-% 
5.'ii:  ••  AB    :P    ::AD.DB:DES 

or  The  Transverse  axis, 

Is  to  its  Parameter, 
As  the  rectangle  of  its  abscisses, 
To  the  square  of  their  ordinate. 

Cor.  3.-If  different  Hyperbolas  have  the  same  Trans' 
verse  axis  the  corresponding  ordinates  in  each,  will  be  to 
each  other  as  their  Conjugate  axes. 

For  as  the  first  and  third  terms  in  the  proposition, 
will  then  be  the  same,  the  second  will  vary  as   the  fourth. 

Fig.  2.  Cor.  4. — Let  APGB  be  the  rectangle  of  the  Transverse 
axis,  and  Parameter  BG,  then  D'HGB  is  the  rectangle  of 
the  absciss  DB  into  the  Parameter,  and  D'IFB  or  the  rec- 
tangle  D'l  .  D'B  is  equal  to  the  square  of  DE. 

For  (Cor.  2.)  AB  :  BG::AD'  .D'B  :  DES 

6.  1     and  Sim.  tri.  AB  :  BG : :  AD'  :  D'l : :  AD' .  D'B  :  D'l' .  D'B, 

and  .-.     AD'  .  D'B  :  D'E^ : :  AD' .  D'B  :  D'P  .  D'B' ; 
D'E2=D'I  .  D'B. 


OF  THE  HYPERBOLA.  69 

ScHOL.  1. — The  square  of  the  ordinate  D'E,  therefore 
(whichequalsD'IFB)is  greater  than  the  rectantjle  of  the  ab- 
sciss D'B  into  the  Parameter,  or  D'HGB,  by  the  rectangle 
IHGF  similar  and  similarly  situated  to  the  whole  rectan- 
gle PB.  It  was  on  account  of  this  excess  in  the  square 
of  the  ordinate,  compared  with  the  rectangle  of  the  ab- 
sciss into  the  Parameter,  that  Apollonius  named  this  sec- 
tion the  Hyperbola. 

ScHOL.  2.  —  If  the  two  axes  are  equal,  that  is  if  the  sec- 
tion is  an  Equilateral  Hi/perbola,  then  the  rectangles  of 
the  abscisses  are  equal  to  the  square  of  the  ordinate,  the 
Equilateral  Hyperbola,  having  the  same  relation  to  a  cir* 
cle,  (which  may  be  called  ^n  Equilateral  Ellipse,)  2iS  any 
other  Hyperbola,  has  to  an  E^llipse. 

ScHOL.  3.  —  From  the  manner  in  which  the  Hyperbolic 
section  is  made,  it  is  evident,  that  as  the  cones  are  more 
or  less  obtuse,  the  Conjugate  axis,  will  be  encreased  or  di- 
minished, while  the  transverse  remains  the  same  ;  and  it 
appears  from  the  3d  Cor.  that  the  curve  will  also  become 
more  or  less  obtuse.  A  section  therefore  may  have  its 
Transverse  axis,  equal  to  the  Conjugate  of  another  sec- 
tion, and  its  Conjugate  may  be  equal  to  the  Transverse 
of  the  other  ;  and  in  this  case  they  are  Conjugate  Hyper- 
bolas, such  are  all  those  considered  in  the  remaining  prop- 
ositions of  this  section,  each  pair  of  opposite  Hyperbolas 
having  the  same  properties. 


PROPOSITION  HI. 

As  the  square  of  the  Transverse  axis, 

Is  to  the  square  of  the  Conjugate, 
So  is  the  sum  of  the  squares  of  the  semi-transverse,  and 

of  the  distance  from  the  centre  to  any  ordinate, 
To  the  square  of  that  ordinate  produced  to  the  Conjugate 

Hyperbola. 

^  10 


70 


OF  THE  HYPERBOLA. 
Thatis,    AB^  lab^y.CA^+CD'  iDer-, 


For  (II  Cor.  3  and  Schol.)  Ca'- 


But, 


CX^::Cd'-Ca'  :de^', 


and 


5> 

or 


C«2 


=Je2  and  De^=Cd''. 


:De-::CA2  :CA-+CD% 
:CAMDe2  CA^+CDS 

CA-'  :Ca2::CA3+CD^  :  Dd^; 

AB2  :a62::CA2+CD2  iDe^. 


In  like  manner  AB^  :  ah^^  :  CA^+CD'^  :  DV^ 
also  ab^  :  AH-^  :  Cfl^+C^i^  -.  dE'\ 

Q.  E. 


I), 


PROPOSITION  IV. 

The  square  of  either  axis,  is  to  the  square  of  the  other,  as 
the  rectangle  of  the  segmentsof  a  line  in  theH^'perbola  par- 
allel to  the  former,  is  to  the  rectangfe  of  the  corresponding 
segments  of  a  line  parallel  to  the  latter. 

f'iS  4.      That  is     eh  .  hg  I  Eh  ,  AG: :  AB^'  :  ab\ 
Or       Eh  .  hG  I  eh  ,  hg'.'.ab''  :  AB^. 

,   ^    For  (1)  ED2  :  eN^::AD  .  DB  :  AN  .  NB, 

and  ED'  :  ED^-cN^ :  :AD.DB  :  AD.DB-AN  .NB. 


Bat 


ED2-eN==ED2-AD2=EA  .  AG, 


And  AD.DB-AN.NB=BD.DN4-NA-BD-DN.NA 


o.  1.   And  since,  BD  .  DN+NA  =  BD  .  DN-fBD  .  NA, 


And  BD— DN  .  NA=BD  .  NA~DN  .  NA, 

Therefore,  AD.DB  — AN  .  NB=DN  .  BD+NA  .  DN 
=  Dtt  ,DN  =  hg  '.  ehf 
ED^  :Eh.  AG:: AD  .  DBr.eh  .  hg, 
And  ED^  :  AD.  DB::EA  ,  hG  l  eh   .  hg. 

5.11.         But    (II.)      EDM  AD.  DB::a62  :  AB% 
EA  .  AG  :  eA  .  hg'.'.ab^  :  AB^. 

Q.  E.  D. 


OF  THE  HYPERBOLA.  71 

Cor.  1 . — Hence  by  equality  of  ratios,  the  rectangles  of  the 
corresponding  segments,  of  lines  parallel  to  the  two  axes, 
mutually  intersecting  each  other  in  the  Hyperbola,  are  to 
each  other  always  in  the  same  ratio;  namely,  the  ratio  of 
the  squares  of  the  two  axes. 

ScHOL. — As  the  square  of  either  semi-axis,  or  of  an  or- 
dinate, is  the  same  as  the  rectangle  of  the  two  halves  of  the 
axis  or  double-ordinate ;  this  proposition  evidently  in- 
cludes Prop.  H  and  HI,  as  it  demonstrates,  that  the  rela- 
tions there  shown  (o  exist  between  the  two  axes,  extend 
to  all  lines  in  the  section  parallel  to  them. 

For      AC^  :  flc^ : :  AD  .  DB  :  DE^   is  the  same  as 
AC    .  CB  :  aC  .  C6:  :AD  .  DB  :  ED  .  DG 
which    may   be   considered   as   included   in   this   propo- 
sition. 


If  the  cutting  plane  move  parallel  to  itself,  until  the 
section  becomes  a  triangle,  the  preceding  proposition 
should  be  applicable  to  the  opposite  triangles  ;  which 
would  also  follow  from  similar  triangles. 

If  the  plane  be  supposed  to  revolve  until  the  section  be- 
comes a  Parabola,  and  the  longersegments  of  the  lines  par- 
allel to  the  transverse  axis  be  considered  equal. 

Then  the  Prop.  AD  .  DB  :  eh  .  A^:  :ED  .  DG  :  Eh  .  AG, 
becomes        AD  :  eA : :  ED  .  DG  :  EA  .  AG, 
The  same  as  is  demonstrated  in  the  Parabola,  Prop.  II, 
Cor.  1. 


PROPOSITION.  V. 


If  lines  parallel  to  the  two  axes,  intersect  each  other, 
without  the  Hyperbola,  the  rectangles  of  their  correspond- 
ing segments  are  to  each  other  as  the  squares  of  the  axes  to 
which  they  are  parallel. 


72  OF  THE  HYPERBOLA. 

Fi?.  5.  That  is,  EH  .  HG  :  eH  .  Hgy.ab'  :  ABS 
or      ell  .  Hg  :  EH  .  HG::AB=^  :  abK 


For  (T,)  cN2  :  ED^ : :  AN  .  NB  :  AD  .  DB, 
And  eN^  :  cN2-ED^•:AN.NB  :  AN.NB— AD.DR. 


But    eN='-ED2=HD2 -ED-^=EH  .HG, 


Also  AN  .  NB=BN  .  AD-hND=BN  .  AD-f  BN  .  DN, 


And   AD  .DB=AD  .  BN  -  ND=BN  .  AD  -  AD  .  DN, 
Therefore  AN  .  NB-AD  .  DB=-BN  .  DN  +  AD  .  DN, 


=DN . BN+AD=DN . BN+Brf 
=DN  .  Nf/=DN  .  D/i. 

Therefore  (eN2=)HD2  :  EH  .  HG: :  AN  .  NB  :  (DN. 
I>n=)eH  .   Hg, 
And        HD2  :  AN  .  NB:  :EH  .  HG  :  eH  .  Hg; 
But  (II,)   HD='(=eN-^)  :  AN  .  NBy.ab'  :  ABs 
EH    .     HG    :  eU  .  Hg  :\ab'  :  AB^. 

Q.  £.  D. 

CoK.  ].  — Since  this  proposition  is  equally  applicable 
to  all  lines  parallel  to  the  axes,  it  follows  by  equality  of 
ratios,  that,  the  rectangles  of  the  segments  of  all  lines 
parallel  to  the  axes,  intersecting  each  other  without  the 
Ellipse,  are  to  each  other  always  in  the  same  ratio  ;  viz. 
that  of  the  squares  of  the  two  axes. 

Cor.  2.  — If  DH  be  supposed  to  move  towards  AP,  the 
rectangle  EH  .  HG  approaches  continually  to  an  equali- 
ty with  the  square  of  AF,  as  its  limit^  and  when  D  coin- 
cides with   A,   AP2  may  be  substituted  for  EH  .  HG. 

That  is,  the  rectangles  of  the  segments  of  lines  parallel 
to  the  axes,  are  to  each  other,  as  the  squares  of  the  inter- 
cepted parts  of  the  tangents  to  the  vertices  of  the  axes. 

f>  -'      Cor.   3. -Since  (IV,)  Eh   .  hG   :  eh  .  hg  Wab^  :  AB% 

^'^-  ^'  and  from  this  EH  .  HG  :  eH .  Hg  :  ab^  :  AB% 

Therefore     VJi  .    hG  :  eh  .  hgWF.R  .  HG  :  f H .  H^. 


OF  THE  HYPERBOLA.  73 

ScHOL.  .  If  the  Hyperbolas  are  Equilateral^  the  ratio 
will  in  all  cases  be  that  of  equality. 
That  is  EH  .  HG=eH  .  Hg. 

Also  if  the  plane  revolve  until  the  section  becomes  a 
Parabola,  and  the  longer  segments  are  considered  equal, 
then  the  rectangles  of  the  double  ordinates  produced,  are 
as  the  external  parts  of  the  diameters. 

That  is  EH  .  HG  is   as  eU. 

As  demonstrated  of  the  Parabola  II,  Cor.    1. 

The  live  preceding  propositions  may  all  be  embraced  in 
the  (oWow'm^  general  proposition.  If  lines  parallel  to  one 
axis  of  the  Hijperhola  intersect  the  other  axis^or  lines  parallel 
to  it^  either  within  or  without  the  section  ;  the  rectanf^les  of 
the  corresponding  segments  will  ahuays  have  to  each  other, 
the  same  ratio. 


PROPOSITION  VI. 

If  a  tang;ent  and  ordinate  be  drawn  to  any  point  of  an 
Hyperbola,  meeting  the  transverse. axis  produced,  the  semi- 
tratisverse  is  a  mean  proportional  between  the  distances 
of  the  two  intersections  from  the  centre, 

That  is,  or  CM  .  CT^CA^  ;  CM  :  CA  :  CT,  are 
continued  proportionals.  Fig 

For  from  the  point  T  draw  any  other  line  TEH  to  cut 
the  curve  in  two  points  E  and  H  ;  from  which  let  fall 
the  perpendiculars  ED  and  HG,  bisect  DG  in  K, 

Then     (1)     AD  .  DB    :  AG     .GB.lDEMGH^, 
And  bysim.  tri.  TD-  :  TG^ ;  iDE^  ;  GH^  ; 

Therefore  AD  .  DB  :  AG  .  GB  ::TD-^  :  TG\ 
But  DB=CB-|-CD=AC+CD=CG-|-DC-AG  =  2CK 

-AG, 
And  GB=CB  +  CG=AC+CG=CG-f  DC  -  AD=2CK 
-AD. 


74  OF  THE  HYPERBOLA. 

.!.    AD  .  2CK-AD  .  AG  :  AG  .  2CK-AG  .  AD:: 
TD3 : TG% 
and  DG  .  2CK  :  (TG^  -TD^or)  DG  .  2TK: :  AD  .  2CK 
5-  17- 16  -AD.  AG  :TDS 

or  2CK  :  2TK: :  AD  .  2CK- AD  .  AG  :  TD^ 

5.  19.  or  AD.2CK  :  AD.  2TK:  :AD.2CK-AD.  AG  :  TD^j 
.'.  AD.2CK  :  AD.2TK::AD.AG:  AD.2TK-.TD% 
5.A&18.  and     CK  :  TK::AD  .  AG  :  AD  .  2TK-TDS 


2.2. &3.  and    CK:TC::AD.  AG:TD2-AD.TJD+TA; 
or      CK  :CT::AD.  AG  .  AT^. 


But  the  limit  of  AD  .  AG,  when  the  line  TH  comes  in- 
to the  position  of  TL,  is  AM*  (See  Parab.  Prop.  IV,)  and 
K,  then  coincides  with  M.  The  proposition  therefore 
becomes, CM  :  CTltAM^  :  AT^  ; 


ThatisCM  :  CT::CM-CA    :  CA  -  CT  , 
5  19.        or      CM  :CT::CM2  +  CA2   :CA2+CT^ 
and     CM  :MT::CM2-1-CA2    :  CM^ -CT% 

or      CM  :  MT:  iCM^  -f  CA^   :  CT-f-CM  .  MT, 
or    CM^  :  CM  .  MT  :  iCM^  ^CA^  :  CM  .  MT4- 
CT  .  MT; 
Hence  CM^  :  CA^ :  :CM  .  MT  :  CT  .  MT, 
^^^        and    CM^  :CA-:CM:CT, 
^^on  •      •'•    CM:CA  ::CA:CT. 

v«rsely.  Q.  E.    D» 


Cor.  l.—CM  :  CT::AM2  :  AT^,  that  is,  the  distances 
from  the  centre  to  the  two  intersections  of  the  tangent  and 
ordinate  with  the  axis,  are  as  the  squares  of  the  distances 
of  the  same  intersections  from  the  vertex. 


CoR.  2. — If  a  tangent  and  ordinate  be  drawn  from  any 
point  in  the  Hyperbola,  and  the  tangent  be  produced  to  cut 
the  conjugate  axia  the  semi-conjugate  is  a  mean  proportion- 
al between  the  distances  of  the  two  intersections  from  the 
centre. 


OF  THE  HYPERBOLA.  75 

For         CD:CL::AC  :CT, 
.:.  CD:CT::€D^  :  CA^ 

.!.  (II,  Cor.)DT  :  CD:  :CD^  -CA^  :  CA»  :  :ED^  :  aC'  ^ 
But  Sim.  tri.  DT  :  CT:  :ED  :  Ct, 
ED  :Q::ED2  :  C^S 
That  is      Cd  :  Ci::Cd^  !  Ca% 
Therefore,  Cd  :  Ca'.'.Ca  :  Ct,  6.19. Cor. 

That  is  Cd  :  Ca  :  Ct,  are  continued  proportionals. 

Cor.  3. — As  the  demonstration  is  applicable  to  a  tan-  ^^?-  '^• 
gent  to  the  curve  at  either  extremity  of  the  double  ordin- 
ate, the  two  tangents  drawn  to  the  two  extremities  of  any 
double  ordinate  to  the  axis,  meet  the  axis  in  the  same 
point,  and  at  equal  angles. 

And,  conversely,  a  line  drawn  through  the  intersection 
of  two  such  tangents,  if  it  bisect  the  angle,  will  also  bisect 
the  double  ordinate  ;  and  if  it  bisect  the  double  ordinate, 
it  will  also  bisect  the  angle,  and  in  both  cases  it  is  the 
axis. 

Cor.  4. —  Also  Kk,  the  tangent  at  the  vertex  is  bisect- 
ed by  the  axis, 

For  TD:TA::DE  :  AK, 

and  TD  :  TA.iDH  :  Ak, 

therefore  DE  :  AK:  :DH  :  AA;; 

but  DE=DH.-.AK=AA:. 

CoR.  5* — If  any  number  of  Hyperbola*  have  the  same 
transverse  axis,  and  an  ordinate  in  one  be  continued  to 
intersect  the  curves  of  all,  the  tangents  from  all  the  points 
of  intersection,  will  meet  the  curve  in  the  same  point; 
for  since  CD,  and  CA,  the  two  first  of  the  continued  pro- 
portionals, remain  the  same  in  all,  the  third  CT  will  be 
the  same. 


CoR.  6. — Two  tangents  to  the  curve  at  the  extremities 
of  any  double  ordinate,  cut  the  conjugate  axis,  at  equal  dis- 
tances from  the  centre. 


76  OF  THE  HYPERBOLA. 

Fig.  8.  Cor.  7. — If  a  perpendicular  from  either  extremity  of 
the  transverse  axis,  be  produced  to  intersect  the  Conju- 
gate Hyperbola,  a  tangent  to  the  curve  at  the  point  of  in- 
tersection, will  pass  through  the  other  extremity  of  the 
transverse  axis. 


Fig.  8.  Cor.  8. — The  following  list  of  proportionals  is  de- 
rived directly  from  this  proposition,  and  its  second  Corol- 
lary, viz. 

That    CT:CA::CA:CD. 

And     Ct  :  Ca  :  :C«  :  Cd. 


(1.)  Then  CT-f  CA  :  CA-CT:  I'^/A+CD  :  CD-CA 
that  is  BT  :  AT::BD  :  AD. 


In  the  same  manner  ht  :  at'.'.bd  I  ad. 

(2.)  Also  CT  :  CT-|-CA::CA  :  CA-j-CD. 
that  is  CT  :  BT::CA:  BD. 

And  Ct:bi::aC  :  bd. 

(3.)  Again  (2.)  CA-CT  :  BD-BT::CT  :  BT, 
That  is       AT:DT::CT:BT. 


And,         at  :  dt::Ct  :  bt. 


(4.)  Therefore  sim.  tri.  AA:  :  DE:  :C^  :  BH, 
and  "     "    Tk:  TE::Tt  :  TH, 


Cor.  9.— When  CD  is  indefinitely  great,  CT  is  indefi- 
nitely  small,  or  the  centre  is  the  point  to  which  the  tan- 
gent approaches  as  a  limit  when  the  absciss  recedes  from 
it. 


OF  THE  HYPERBOLA.  77 


PROPOSITJON   VII. 

The  rectangle  of  the  Focal  distances  from  the  vertices, 
is  equal  to  the  rectangle  of  one  fourth  the  parameter,  and 
the  transverse  axis. 

That  is  AF  .  FB=iP  .  AB.  Fig.  8. 

For  (II  Cor.  2.)  AF  .  FB  :  FE^ : :  AB  :  P, 

therefore        AF  .  FB  :  FE^ : :  AB  .  AF  :  P  .  AF, 
and  AF  .  FB  :  AB  .  AF.iFE^  :  P  .  AF ; 

But  (19)        FE  =  iP.-.FE-"=iPS 

AF  .  FB  :  AF  .  AB  :  {P^ :  :P  .  AF, 
FB  :  AB::{P  :  AF, 
and  FB  .  FA^^P  :  AB. 

Q.  E.  D. 

Cor.  1.— FB  :  ABii^P  :  AF. 

ScHOL. — If,  as  in  the  Parabola,  FB  and  AB,  be  consid- 
ered equal,  then  ^P=AF  as  demonstrated  in  the  V.  Prop, 
ef  Parabola. 

CoR.  2.— Since  FB  .  FA=iP  .  AB,  and  FB  is  greater 
than  AB, 

FA  is  less  than  {P. 
In  the  Parabola,  FA=iP,  in  the  Ellipse  FA>iP. 

CoR.  3.— AF  .  FB=CaS  or  the  rectangle  of  the 
Focal  distances,  equals  the  square  of  the  semi-conju- 
gate. 

For  since  (19.)  P.  AB=a62.-,iP  .  AB=aC2=AF.  FB. 
or  AF  :  aC:  :  aC  :  FB. 


Cor.  4.— AF.FB(=flC2)  =  |a^2.  or  iAB.iP=AC.FE ;  f'^g-  10. 
or        AF:  FE::AC  :FB. 

11 


4> 


70  OF  THE  HYPERBOLA. 

^iS'  8.      Cor.  5.--(VI  Cor.  1.)  aC^=C<  .  C^=C/  .  DK, 
but  (VI.  Cor.  7,  4.)C<  .   DE  =  AA:  .  BH, 
^  .-.     A^  .  BH=flC-^=CA  .  |P=AF.  FB. 

Cor.  6. — Therefore  H/A:,  is  a  right  an^le. 
for  (Cor.  5)  AA;  :  AF::FB  :  BH, 
.*.  the  triangles  A^and  BHF  are  similar  ; 
.-.    -  angle  BH/=A/A:  and  B/H  =  A^/; 
.»*.  also  Hfk  is  a  right  angle. 

The  same  may  be  proved,  in  reference  to  the  other 
Focus,  therefore  a  circle  described  on  HA:,  as  a  diameter, 
will  pass  through  the  two  Foci. 


PROPOSITION  VIII. 

The  square  of  the  distance  of  the  Focus  from  the  cen- 
tre,   is  equal   to    the   sum    of  the   squares    of  the  semi- 
axes. 
**•§:•  »•  That  is  CF^  =CA2+C<r2, 

For  (III  Cor.  1.)  CA^    :  Ca^ :  iCF^ -CA=^   :  FE», 
6.22.  and  (19)  CA^    :  Ca^y.Ca^   :  FES(=iPS) 

Therefore  CA'^ -CF^^CaS 

and  CF2=CA=-fCa^ 

Q.  £.  D. 

Tis-  9.      Cor.  1  .—Hence  FF^  =  AB^  -|-a6« . 

CoR.  2. — The  two  semi-axes,  and  the  Focal  distance 
from  the  centre,  are  the  sides  of  a  right  angled  triangle, 
in  which  the  hypotenuse,  is  equal  to  the  distance  of  the 
Focus   from  the  centre. 

For    CA2-|-Ca-=Aa% 

and     CA2  4.Ca='=CF%.-.Aa=CF. 

Cor.  3. — Hence  the  distances  of  all  the  Foci,  of  any 
pair  of  Conjugate  Hyperbolas  are  equal,  being  each  equal 
to  the  sum  of  the  squares  of  the  semi-axes. 


OF  THE  HYPERBOLA.  79 

Cor.  4. — The  square  of  the  Focal  distance  from  the 
ntre,  is  equal  to  the  rectangle  of  the  semi-transverse, 
and  the  sum  of  the  semi-transverse  and  the  semi-param- 
eter. 


CO 


That  is     FC-=CA  .  CA+FE, 
For      FC^=CAMh^~^and  aC^=(19)  AC  .  FE, 
Therefore  FC-^=CA34-AC  .  F%  =  AC.  aU+T^, 


and.-.     AC:FC::FC  :  AC-fFE.  6.17. 


PROPOSITION  IX. 


The  difference  between  two  lines  drawn  from  the  Foci 
to  meet  at  any  point  in  the  curve,  is  equal  to  the  trans- 
verse axis. 

That  is    /E-FE=AB.  Fig.  lo- 

For,  draw  AG  parallel  and  equal  to  Ca  the  semi-conju- 
gate; and  join  CG  meeting  the  ordinate  DE  in  H  ;  also 
take  CI  a  fourth  proportional  to  CA,  CF,  CD  ; 

Then(n,CoR.l.)  CA-^  :  AG^ :  iCD^ -CA=  :  DE^  6.22.it 

And,  (sim.  tri.)  CA^  :  AG^ ;  iCD^  _CA^  :  DH^  -  AG^     ^-  '^• 
consequently  DE^  =DH^-  -  AG^  =D}1^  -Ca\ 


Also       FD,  is  the  difference  between  CF,  and  CD, 
And      FD2=CF2-2CF.CD+CD% 
FE2=FD  =+DES 
Therefore   FE^  =CF2  -  Ca'  -  2CF  .  CD-f-CD^  +DH^ 

But(Vlll.)  CF2-Ca2=CA2; 

and  (Hyp.)  2CF  .  CD  =  2CA  .  CI  ; 

Therefore    FE2=CA2-2CA  .  CI-|-CD='+DH^ 


2.  4, 


>.4.Cor.     jjj  the  same  manner  it  is  found  that,  /'E=CA-fCI=BI  ; 


80  OF  THE  HYPERBOLA. 

Again  (Hyp.)  AC^  :  CD^* :  :CF^(or  CA^-f  AG^)  :  Cl^  ; 
And  CA2  :  CD»  :  iCA^-t-AG^  :  CD^  +  DH^ 

CI==CD2+DH2=(  H% 

ConsequentIy.FE2=CA2-2AC  .  Cl  +  CP. 

And    the   side   of  this   square    is   FE=CI  — CA=AI; 
lanner  it  is  found  that,  fE  = 
/E-FE=BI-AI=AB. 

Q.  E.  D. 

Cor.  1. — The  sum  of  the  semi-transverse,  and  the 
distance  from  the  Focus  to  any  point  in  the  curve,  is 
a  fourth  proportional  to  the  semi-transverse,  the  distance 
from  the  centre  to  the  Focus,  and  the  distance  from  the 
centre  to  the  ordinate  to  that  point. 

That  is  CA  :  CF:  :CD  :  CI.  (or  CA+FE). 

ScHOL. — From  this  proposition  is  derived  the  com- 
mon method  of  describing  this  curve  mechanically,  by 
points. 

F'g-  9       In  the  Transverse  take  the  Foci  F,yand  any  point  I. 

^°  ^  ■  Then  with  the  radii  AI,  Bl  and  centres  F,  /;  describe 
arcs  intersecting  in  f,  which  will  be  a  point  in  the  curve. 
In  like  manner,  assuming  other  points  I,  as  many  other 
points  will  be  found  in  the  curve.  Then  with  a  steady 
hand,  the  curve  line  may  be  drawn  through  all  the  points 
of  intersection  c. 


PROPOSITION  X. 

If  there  be  any  tangent,  and  two  lines  drawn  from  the 
Foci  to  the  point  of  contact  ;  these  two  lines  will  make 
equal  angks  with  the  tangent. 

Fig.  10.      For  draw  the  ordinate  DE,  and  fe  ;  parallel  to  FE, 
Then(IXCor.  1)CA  :  CD::CF  :  CA-{-FE; 
and  (VI)       CA  :CD;:CT  :  CA, 
therefore      CT  :  CF:  :CA  :  CA  +  FE, 
2.  4.  and  TF  :  T/:  :FE  :  2CA-f  FE,(or/E)— (IX); 

But  (sim.  tri.)    TF  :  T/:  :FE  :/e, 

therefore     /E=/e  and  the  angle  e=the  angle/Ee  ; 
but  because     FEis  parallel  tofe,  the  angle  e=the    an- 
gle FET  ; 
therefore     The  angle  FET  =  the  angle/Ee. 

Q,  E.  D. 


OF  THE  HYPERBOLA.  81 

ScuoL. — As  opticians  find  that  the  angle  of  incidence 
is  equal  to  the  angle  of  reflection,  it  appears  from  this 
proposition,  that  rays  of  light  issuing  from  one  focus,  and 
meeting  the  curve  in  every  point,  will  be  reflected  into 
lines  drawn  through  those  points  from  the  other  focus,  so 
the  ray/E  is  reflected  into  FE,  and  this  is  the  reason  why 
the  points  F,/,  are  called  the  foci  or  burning  points. 

Cor.  1. — Hence  the  JVbrma/,  EO,  or  the  line  drawn 
perpendicular  to  the  tangent,  from  the  point  of  contact, 
bisects  the  angle  made  by  the  two  lines,  drawn  from  the 
Foci. 

For  since  the  Normal^  makes  equal  angles  with  the  tan- 
gent, .     3 
OEP=OEF,  or  FE/ is  bisected  by  EO. 

Cor.  2.— Consequently  FE  :/E::FO  :/0;  6-3. 

and  FE+/*E  :/E-FE::FO+/0  :/0-F0;   ^- ^• 

That  is  (IX)  2CI  :  2CA:  :2C0  :  F/, 
therefore      CA   :    CI::CF  :  CO. 

CoR.  3.-Therefore  CA^  :  CA  .  CI:  iCF^  :  CF  .  CO, 
and      CA2  :  CF2 ;  :(CA  .  CI=)  CF  .  CD  :  CF  .  CO, 
.-.      CA2  :  CF^::CD:CO. 

Cor.  4.-Therefore  CF^  -CA^  :  CF^ :  :CO-CD  :  CO. 

that  is  (VIII)        C«-^    :CF^::DO:CO. 

CoR.5.— CA»  :  CF2  -CA^ :  :CD  :  CO -CD. 
that  is     CA^  :  Ca=::CD  :  DO. 

ScHOL.  — When  the  Hyperbola  is  Equilateral,  CD  = 
DO  ;  and  CE=EO  ;  or  (Fig.  13.)  CE=EI. 

Cor.  6.-Hence  (19)  AB  :  P:  :CD  :  DO,  6.  20. 

and  AC  :iP::CD:DO,  ^^^'^•*' 

AC.DO=^P.CD 
And  if,  as  in   the    Parabola,  AC  and  DC,  be  supposed 
equal,  then  DO,  the  sub-normal,  equals  half  the  Parame- 
ter. 


82  OF  THE  HYPERBOLA. 

Cor.  7.-2CA  :  2CI::/F:  2C0::CF  :  CO, 
2CA:  2CI-2CA::CF:CO-CF. 
that  is  (IX)  2C  A  :  2FE  : :  CF  :  OF  ; 

or  CA:FE  ::CF:OF. 


PROPOSITION  XI. 

If  a  line  be  drawn  from  either  Focus,  perpendicular  to 
a  tangent  to  any  point  of  the  curve,  the  distance  of  their 
intersections  from  the  centre  will  be  equal  to  the  Semi- 
transverse  axis. 

Fig.  n.       That  is  CP  and  Cp  each=CA  or  CB. 

For  through,  the  point  of  contact  E  draw  FE,  and/E 
meeting  FP  produced  in  G,  then  the  angle  GEP=angle 
FEP,  and  the  angles  at  P  being  right  and  the  side  PE  be- 
ing common,  the  two  triangles  GEP,  FEP,  are  equal  in 
all  respects,  and  so  GE  =  FE,  and  GP  =  FP.  Therefore, 
since  FP  =  ^FG,  and  FC  =  |F/,  and  the  angle  at  F  com- 
mon, the  side  CP  will  be  =  2/G  or  ^AB,  that  is  CP=CA 
or  CB. 

And  in  the  same  manner  C»=CA  or  CB. 

Q.  E.  D. 

CoR.l.— A  circle  described  on  the  Transverse  axis,  will 
pass  through  the  points  of  intersection  P  and  p. 

Cor.  2. —  If  from  the  intersections  of  a  tangent  with  a 
circle,  described  on  the  transverse  axis,  perpendiculars  be 
drawn,  they  will  pass  through  the  Foci. 

Cor.  3.  —The  distances  of  the  Foci  from  (he  point  of 
contact,  are  to  each  other  as  the  perpendiculars. 

For  the  triangles  EFE,fpE  are  similar  ; 
Therefore  FE    :/E::FP  :  fp. 

CoR.  4.-If  PF  and  Cp  be  produced,  they  will  meet  in 
the  circumference  of  the  circle  in  K,  of  which  pK  will  be 
the  semi-diameter,  for/PK  is  a  right  angle. 


OF  THE  HYPERBOLA.  gg 

also  FKC,  C/j/'are  similar  and  equal  trians;les ; 
Therefore  ?F  .  pf=?F    .  FK=AF  .  FB=CG^  (VII. 
Cor.  5.) 


PROPOSITION  XII. 

The  distance  of  the  Focus  from  any  point  in  the  curve, 
is  equal  to  the  ordinate  to  that  point,  continued    until  it 
meets  the  Focal  tangent, 
thatisFA=AL,  FB=Bz,  and  generally  FM=DG,  *^ig- 12. 

for  (19)  FE=iP, 

and  (VI  Cor.  2)  FE  ."C/=Ca2=FE.  AC, 
therefore  C/  =  AC. 

Again  (VI)  CF  :  AC::CA  :  CT, 
and         AC  :  AF::CT  :  AT, 
Sim.  tri.      AT  :  TC::AL  :  C/,  =  CA; 

AF  :  AC::AL  :  AC,  , 

therefore    AF  =  AL. 

That  is,  the  distance  from  the  vertex  to'  the  Focus,  is 
equal  to  the  tangent  to  the  vertex,  intercepted  by  the 
Focal  tangent. 

(VI  Cor.  7,)     AF  :  FB::AT  :  TB, 
.-.Sim  tri.       AT  :  TB: :  AL  (  =  AF)  :  BZ, 
therefore      AF  :  FB: :  AF  :  BZ, 
BZ=BF. 

rVI.)  AC2  =CF  . CT=CF  .  CF-FT=CF='-CF .  FT 

'  .-.  CF2-AC2(=:Ca^)  =  CF.  FT, 
^im.  tri.  TF  :  FE :  :TC  ;  Ct=AC, 
but    TC  :  AC:  :TC  .  CF  :  AC  .  CF, 
.  .     TC  .  CF  (  =  AC2)  :  AC  .  CF:  :CA  :  CF, 
thatis    TF:FE::  (AC  :  CF):  :TA  :  AL:  :TD  :DG  : 
but  (IX)  CT  :  AC :  :CD  :  FM-fCA, 
and       CD-CT  :FM-|-CA-CA::CT  :  CA, 
thatis    TD:FM::CT:  AC; 

.'.      TD:DG::TD:  FM.'.DG=FM. 

Q.  E.  D. 


g4  OF  THE  HYPERBOLA. 

Cor. — If  through  T,  a  line  be  drawn  perpendicular 
to  TA,  it  is  called  the  directrix. 

The  distance  of  any  point  in  the  curve,  from  the  direc- 
trix as  MN  is  equal  to  TD  which  has  a  constant  ratio  to 
DG  or  FM,  the  distance  of  the  point  from  the  Focus.  In 
the  Parabola  it  is  a  ratio  of  equality. 


PROPOSITION  XIII. 

If  tangents  and  ordinates  be  drawn  from  the  extremities 
of  any  two  conjugate  diameters,  meeting  the  Transverse 
axis  produced,  the  distances  of  these  intersections  from 
the  centre  are  reciprocally  proportional. 

Fig.  13.     That  is    CD  :  Cd:  :Ct  :  CT. 

For  (VI)    CD  :CA::CA  :  CT, 
and  „    Cd  :  CA::CA  :  C^ 
CD  :Cd::Ct  :  CT. 

Q.  E.  D. 

CoR.  1. — Hence  the  distance  from  the  centre,  to  the 
ordinate,  at  the  extremity  of  one  diameter,  is  a  mean  pro- 
portional between  the  distances  of  the  intersections  of 
the  tangent  and  ordinate  drawn  from  the  extremity  of  the 
other. 

For  since      CD  :  Cd:  :Ct  :  CT, 

andbysim.tri.  C(/:  DT::C/  :  CT, 

CD  iCdr.Cd:  DT. 

In  hke  manner  Cd  :  CD:  :CD  :  dt. 

Cor  2.  — The  ordinates,  also,  are  reciprocally  as  their 
distances  from  the  centre. 

For  Cor,  1.  and  sim.  tri.     DE  :  der.Cd:  CD. 


OF  THE  HYPERBOLA.  85 

•  Cor.  3. — Therefore  the  ordi nates  are,  as  the  distan- 
ces of  the  intersections  of  their  tangents  from  the  centre, 
or  (Cor.  1,)  DE  idey.CT  :Ct, 

Cor.  4. -Also  by  (Cor.  2,)  The  rectangles  DE  .  CD 
=de  .  Cd', 
therefore,  the  triangle  CDE=Cc?f. 


PROPOSITION  XIV. 

If  ordinates  to  the  Transverse  axis,  be  drawn  from  the 
extremities  of  any  two  conjugate  diameters,  the  difference 
between  their  squares  is  equal  to  the  square  of  the  Semi- 
Conjugate  axis,  and  the  difference  between  the  squares 
of  two  ordinates,  drawn  from  the  same  extremities  to  the 
Conjugate  axis,  is  equal  to  the  square  of  the  Semi-Trans- 
verse. 

That  is,         de^  -I)E^=CaK  F'g-  13. 

And  ED'^—ed"'=CA\* 

That  is  CD'—Cd^-==CAK 

For  (VI)       CD  :CA::CA:CT, 
.:.  CD  :CA::AD  :  \T  ; 

And  CD:DB::AD:DT, 

...(XIII.  1)  CD  .DT  (  =  Cc/-)=AD.DB=CD»-CAS 

Cc/2=CD--CA=^; 
And  CD2-CGf2=CA2;  m 

That  is  ED'^-e'd'=CA\  I 

In  the  same  manner  c/c---DE2=Ca2.  l 

q,  E.  D. 

CoR.  1.— Hence  CD .  DT=AD.DB=:CJ^  =CD^  -CA^; 

CoR.  2.— Hence     CD^  -  Cd^  =CA2  ; 

and         Cc/'2-CD'2=((ie2-DE2)  Ca-  ; 

*The  ordinates  to  the  Conjugate  axis,  from  E  and  c,  are  not  drawn  in 
the  figure  (13),  but  they  are  evidently  equal  to  CD  and  Cd. 

12 


dQ  OF  THE  HYPERBOLA. 

Cor.  3.— Since     CA«  :Ca^ :  :(AD.DB=)  CJ^jDE*, 
.-.       CA:Ca::C</:DE. 
and       CA  :  Co  I :  CD  :  de. 
In  Equilateral  Hyperbolas   CD=rfe  and  C£i=DE. 


PROPOSITION  XV. 

If  from  the  /extremity  of  any  diameter  a  perpendicular 
be  drawn  to  its  conjugate,  the  rectangle  of  this  perpendic- 
ular, and  the  part  of  it  intercepted  by  the  Transverse  Axis, 
is  equal  to  the  square  of  the  Semi-Conjugate  axis. 

^''&-i3.     Thai  is        EP.EI=Ca«. 

Draw  Cy  parallel  to  EP,  the  triangle  EDI,  and  CyT  are 

similar  : 
Therefore  CT  :  Cy:  :E1 :  ID, 

That  is  CT:EP::EI  :  ID, 

Therefore  EP  .  EI=CT  .  ID=C^' .  DE=Ca^ 

Q.  E.  D. 


PROPOSITION  XVI. 

All  parallelograms  described  between  four  Conjugate 
Hyperbolas,  whose  sides  are  parallel  to  two  conjugate  di- 
ameters, are  equal  to  each  other,  being  each  equal  to  the 
rectangle  of  the  two  axes. 


F15.  13.     That  is 

KQI 

^S= 

AP. 

ab. 

For(XIV.Cor.3)CA 

:Ca 

::CD 

^:de; 

Or 

CA 

:CD 

::Ca 

:de. 

But  (VI) 

CA 

:CD 

::CT:CA, 

Ca: 

:de:: 

CT: 

;CA, 

Also,  sim.  tri. 

Ce  : 

de:: 

CT: 

%» 

Therefore 

Ca: 

Ce: 

:Cy: 

CA; 

And 

Ca. 

CA= 

=Ce. 

%(  = 

=EP.) 

4Ca, 

,  CA,  or  AB  .  ab: 

=r4Ce  . 

EP,  01 

r  QKSN. 

«. 

E.D. 

OF  THE  HYPERBOLA  gj 


PROPOSITION  XVII. 

The  difference  between  the  squares  of  every  pair  of 
conjugate  diameters,  is  equal  to  the  same  constant  quanti- 
ty, viz.  the  difference  between  the  squares  of  the  two  axes. 

That  is,  AB^ --ab^  ^IIG^ -eg^.  Fi-.   U 

For  CE2-C62=CD^-CJ^4-DE2-De2;         i- ^' 

But  (XIV)  CD'-Cd'=AC';  amd  de' -DE'=Ca'; 
therefore      CE»— Cc^rrrAC^— Ca^, 

and  EG^-eg^=AB^-abK 

Q.  £.  D. 

ScHOL. — In  Equilateral  Hyperbolas,  all  conjugate  diam- 
eters are  equal  to  each  other. 


PROPOSITION  XVIII. 

If  the  extremities  of  the  transverse  and  conjugate  axes 
be  connected,  lines  drawn  through  the  centre,  parallel  to 
the  connecting  line,  are  the  Assymptotes,  and  may  be  con- 
sidered equal  ;  and  as  the  extremities  of  two  conjugate 
diameters,  recede  from  the  vertex,  the  diameters  approach 
to  the  Assymptotes,  and  to  a  ratio  of  equality. 

For  Aa=aB=Bb=^bA.' .AaBb,  is  a  parallelogram,  PL  VIII. 

also  the  angles  GCB=aAB=^AB=LCB,  ^'-'  ^^^ 

CG,  CL,  are  parallel  to  Aa  and  Ah. 

That  both  the  Conjugate  diameters  increase  in  length,  or 
approach  to  the  Assymptotes,  when  either  does,  and  also 
that  they  approach  to  a  ratio  of  equality,  is  evident  from 
the  last  Proposition. 

Q,  E.  D, 


3g  OF  THE  HYPERBOLA. 


PROPOSITION  XIX. 


If  from  any  point  of  the  Hyperbola,  a  Normal  be  drawn 
and  produced  to  cut  both  the  axes,  the  part  intercepted  by 
the  Conjugate  shall  be  to  that  intercepted  by  the  Trans- 
verse, as  the  square  of  the  Transverse  is  to  the  square  of 
the  Conjugate. 

PLXVI.8.     That  is  EI  :  EO: :  AB^  :  ah\ 


For  AB=»  :a6*::CD  :  DO, 

::EI  :  EO. 


Cor.  EF  :  E/::OF:  Of, 


Q.  £.  Z). 


PROPOSITOIN  XX. 

If  a  tangent  and  ordinate  be  drawn  from  the  extremi- 
ties of  the  transverse  axis,  and  of  any  other  diameter,  and 
be  produced  to  meet  that  diameter  and  transverse  axis,  th« 
triangles  formed  respectively  by  these  lines,  are  equal. 

IS-  17-     That  is,    CAE=CMT,  and  AGT=MGE, 
also      CHN=CDM,  and  AHE=MDT. 

1.  For  CDM  and  CAE,  also  CHA  and  GMT  are 

similar. 
.i.  CH  :  CM::CA  :  CT, 

::CD  :  CA, 
::CM  :  CE; 
therefore  DH,  AxM  and  TE  are  parallel; 

AME  =  AMT,  and  CAE=CMT. 
6.  15.  2.    Also  taking  CEGT  from  each,  AGT=MGE. 

3.  "    adding  MDA^MHA  and  MGA,  to  each,  MDT= 

4.  therefore     CDM=CHA.  [AHE. 

Q.  jE.  a 

CoR.l.— AME4-AMD=MDAE=MDT  ; 
AMH-hAMT=AHi\fT=-AEH ; 
MDAE=AHMT. 


OF  THE  HYPERBOLA.  g^ 

Lor.  2.         QID=AEFI.  ^^s   14 

For  sira.tri.  CA  ;  AE:  :CD  :  DM:  :CD-f-t:A  :  DM-^AE; 
and  CA  :  AE::Cl  :  IF::C1+CA  :  IF+AE ; 

Again       AI ,  IB=AI .  IC+CA  ; 

and  AD  .  DB  =  AD  .  DC+CA ; 


therefore  AI. IB  :AD.DB::AI.IC-hCA:  AD.  DC+CA* 


AI.AE+IF:AD.AE4-DiM, 
2AEFI:  2AEMD, 
AEFI  :  AEMD. 
But  sira.  tri.  QID  :  MDTiiQl^    :   JV1D% 

::Al  .  IB  :  AD  .  DB, 
::AEFI  :  AEMD; 
But  (Cor.  1.)     MDT=AEMD  .  .  QID=AEFI. 

Cor.  3.  QLF=LMTD. 

For  from  CLD  take  CAE  and  CMT, 
then    LDTM=LDAE=AEFI  -LDIF=DQI-LDIF 
=LQF. 


PROPOSITION  xxr. 

The  square  of  any  diameter. 
Is  to  the  square  of  its  conjugate, 
As  the  rectangle  of  its  abscisses. 
Is  to  the  square  of  their  ordinate. 

That  is         NM2  :  VK^iiNL  .  LM  .  LQ-.  Fig.  i4 

For  (XX.  Cor.  3.)         LQF=LD TM  ; 

Now  suppose  QLD  to  move  parallel  to  itself,  towards 
KC,  it  will  first  coincide  with  the  tangent  TM,  and  then 
both  LQF  and  LDTM  will  become  equal  to  nothing.  As  it 
passes  the  tangent  grid  comes  into  the  situation  KC,  KCO 
corresponds  to  QLF,  and  IMC  to  LTDM ; 
therefore  KCO  =  TMC.* 

*The  transfer  of  the  relation  of  LQF  and  LDTM,  to  KCO  and  TMC, 
is  founded  on  the  analosy  between  Prop.  II  and  III  The  equaliiy  ot  KCO 
and  TMC,  may  be  rigorously  deduced  from  Prop.  III.— See  Emerson's 
Con.  Sect.  29  'tnd  36. 

Also     agC=TMC+C%;       ...     aK:-CTM. 


90  OF  THE  HYPERBOLA. 

6.  20.  But  fim.  tri.  CMT  :  CDL:  :CM=»  :  CLS 

and         CMT  :  CDL-CMT:  :CM=^  rCL^-CM^, 
that  is       CMT  :  LDTM.'.CM^  :  NL .  LM ; 
But  Sim.  tri,  CK^  :  LQ^::CKO  :  LQF, 

::CMT:  LDlM, 
::CM^  :  NL.LM, 
therefore  CK^   :  CM^.iLQ^'   :  NL  .  LM ; 
and         NM^  :  VK^::NL.LM  :  LQ^ 

Q.  E.  D. 

ScHOL. — The  preceding  proposition  shows,  that  any  di- 
ameter of  a  Hyperbola,  and  its  rectangles,  have  the  same  re- 
lation to  its  conjugate,  and  ordinates,  as  have  been  demon- 
strated in  the  case  of  the  Axes.  All  the  properties  therefore, 
which  have  been  deduced  from  this  relation,  in  the  former 
case  in  reference  to  lines  intersecting  each  other,  parallel  to 
the  axes,  and  to  tangents  intersecting  these  lines  and  the 
curve,  may  be  applied  to  any  conjugate  diameters.  The 
first  six  propositions,  therefore,  of  this  chapter,  with  their 
corollaries,  are  particular  truths  which  may  safely  be  gen- 
eralized by  applying  them  to  any  two  conjugate  diameters, 
as  well  as  to  the  two  axes.  It  is  not  thought  necessary 
to  state  these  general  propositions,  since  they  correspond 
to  those  concerning  the  axes,  in  precisely  the  same  man- 
ner as  in  the  Ellipse. 

The  relations  of  the  Hyperbola  to  its  Asymptotes,  form 
a  peculiar  class  of  properties,  to  which  there  is  nothing 
corresponding  in  the  other  sections  ; — a  few  of  them  will 
be  stated. 


PROPOSITIOxN  XXII. 

Jf  a  double  ordinate    be  produced  to  meet  the  Asymp- 
totes, the  rectangle  of  its   segments    will    be  equal  to  the 
square  of  the  semi-axis  to  which  it  is  parallel. 
Fig.  15.      That  is  HE  .  EK=He  .  fK=Ca^ 


OF  THE  HYPERBOLA,  9 1 

and     he  .  tk-^hE  .  EAr^CA^. 
For  CA^  :  )ALs  =  )Ca2::CD2  :  DH% 

And  (II.  Cor.)  CA"  :  Ca^ :  iCD^-CA*  :  DE-', 
Therefore  CA^  :  Ca^iiCA^  :  DH^ -DE2=HE  .  EK,      5  19. 
*•       CA-=HE.EK. 
Again    CA=^  rCa^iiCD'  :  DHs 
And  (III)  CA^  :Ca2::CA2+CD«  :  DeS 

CA2  :Ca2::CA2  :  De^-DH3=Hc  .eK, 
Therefore  Ca2=He  .  eK. 

In  like  manner,  in  the  Conjugate  Hyperbolas, 
he  .  eh  and  AE  .  EA:=CA^ 

Q.  E.  D. 

ScHOL. — In  Equilateral  Hyperbolas  He  .  cK=Ae  .  ek. 

Cor  1.— Since  HE  .  EK  =  He  .  cK, 
EH  :eH::eK:EK. 

Cor.  2— sim.  tri.  EA  :  EH::EA:  :  EK,  • 
AE  .  EK=HE  .  EA:. 

Cor.  3.— HE=EK,  and  EG=E^,  and  He=Ke. 

Cor.  4.~Since  HE  .  EK=L  A^^LA  .  AI, 

G£  .  EK=PA  .  Al.  6.  14. 

Cor.  5.— em,  AS,  EK,  being  parallel  to  HC,  Fig.  14. 

GE  .  EM=PA  .  AS=NE  .  EK, 
and  GEMC =PASC  =NEKC. 

Cor.  6. —  Since  all  rectangles  between  the  cur^e 
and  Asymptotes,  as  CG,  GE,  are  equal,  therefore  GE,  is 
reciprocally  as  CG;  and  if  CG  be  increased  indefinitely 
GE,  will  be  indefinitely  diminished  ;  that  is  the  curve  con- 
tinually approaches  the  Asymptote,  but  never  meets  it.  It 
is  considered  a  tangent  to  the  curve  at  an  infinite,  distance. 

CoR.  7. — If  CI  :Cg  :  Cn,  be  continued  proportionals, 
IK  :  ^E  :  rm,  parallel  to  the  other  Asymptotes,  are  contin- 
ued proportionals,  decreasing  ;  for  they  are  reciprocally  as 
C/,  C^,  Cn,  being  sides  of  equal,  and  equiangular  paral- 
lelograms. 


92  OF  THE  HYPERBOLA. 

Cor.  8.— CAE=PAEN=SAEK. 
For  CPAS«=CNEK    .PI=1K, 

therefore  PaEN=SaEK. 

Again  from  CAEK  take  CAS.  and  AEK. 
and     CaE=SAEK=PAEN. 

Cob.  9. — If  a  cutting  plane  be  supposed  to  pass  througk 
PI  VIII  ^^^  vertex  of  the  cone,  parallel  to  the  plane  which  forms 
Fig.  19.  the  Hyperbolic  section,  forming  the  triangular  section 
VCD ;  and  if  two  planes  VGC,  VGD  be  supposed  to 
touch  the  convex  surface  of  the  cone,  in  the  sides  of  this 
triangle,  then  the  intersections  of  these  planes  and  the  plane 
of  the  Hyperbola,  will  be  the  Asymptotes  of  the  Hyper- 
bola. 

That  is,   rK,    rL,  are    Asymptotes  to    the  Hyperbolic 
section,  HAiI;  for  HK,  evidently  equals  LI,  and  M=/t,.-» 
KH    HL=K1  .  XL.  also  kh  .  hl=:ki .  il. 
3.  36.      Also  KH  .  HL=KC' =kc^  =kh  .  hi.  &c. 


OF  THE  CURVATURE  OF  CONIC  SECTIONS. 


It  is  evident  that  the  different  parts  of  the  curve  of  any 
Conic  Section,  have  different  degress  of  curvature.  At 
the  vertices  of  the  Transverse  Axis,  for  instance,  they  are 
more  curved  than  in  other  parts,  and  the  curvature  evi- 
dently decreases  with  the  distances  from  these  vertices. 
A  person,  unacquainted  with  the  doctrine  of  the  Curva- 
ture of  Curves^  might  naturally  describe  the  difference  be- 
tween the  curvature  of  an  Ellipse,  at  the  vertices  of  its 
Transverse  and  of  its  Conjugate  axis,  by  saying  that  a  cir- 
cle, which  should  have  the  same  curvature  as  the  Ellipse, 
at  the  vertex  of  its  Transverse  Axis,  or  which  should  co- 
incide with  the  curve  at  that  point,  would  be  smaller  than 
the  circle  which  would  coincide  with  it,  at  the  vertex  of 
the  Conjugate  Axis.  This  obvious  method  of  explaining 
the  subject  is  substantially  the  same  as  that  given  by  Math- 
ematicians, and  is  capable  of  the  precision  of  mathematical 
demonstration  to  a  greater  extent  than  might  at  first  be  sup- 
posed. Although  no  definite  portion  of  a  circle  and  Co- 
nic Section,  can  possibly  coincide,  since  the  nature  of  the 
curves  and  their  equations  are  essentially  different,  yet,  at 
their  point  of  contact,  they  may  have  the  same  Curvature, 
and  are  often  said  to  coincide.* 

Def.  1. —  The  Curvature  of  a  line,  is  its  continued  devi- 
ation from  a  straight  line;  and  the  degree  of  its  curvature  is 
measured  by  the  perpendicular  distance  of  any  point  in  the 
curve,  from  the  straight  line,  at  a  given  distance  from  their 
point  of  contact  or  concourse. 

Thus,  DE,  DF,  which  are  called  suhtenses  of  the  an-  ^J?- 
gles  DCE,  DCF,  being  the  distances  of  the  points  E  and 
F  in  the   curves,   from  the  straight  line  AD,  are  the  meas- 
ures of  the  curvature  of  the  two  circles,  or,  more  accurate- 

*  Bridge'?  Conic  Sections,  pp.  7.5, 80. 
13 


jl4  Of"  THE  CURVATURE  OF  CGMC  SECTIO.NS. 

Zy,  if  the  line  DEF,  were  to  move  parallel  to  itself  towards 
C.  unti'  it  should  coincide  with  CG,  the  limiting  ratio,  of 
DE  to  DF,  would  be  the  ratio  of  the  curvatures  of  the  two 
circles,  at  that  point.  This  limiting  ratio  of  the  subtenses, 
in  many  cases,  is  easily  obtained. 

*''o  1  Lemma.— If  from  (' and  E,  any  two  points  in  the  cir- 
cumference of  the  circleCNE,  straight  lines  be  drawn 
to  any  other  point,  V,  and  a  straight  line  CD,  touching  the 
circle  in  C  ;  and  if  from  E,  ED  and  Ew  be  drawn  parallel 
toCVand  CD,  also  the  chord  CE, — then,  if  E  be  suppo- 
sed to  move  in  the  curve  towards  C,  the  tangent  CD, 
the  chord  CE,  and  the  ordinate  Ev,  approach  continually 
to  a  ratio  of  equality,  as  xheir  limiting  ratio-,  they  are  there- 
fore ultimately  equal. 
3.  32  For,  since  the  angles  DCE,=EVC,  and  CV,  DE,  are 
paral.  therefore  by  sim.  tri.  CE  :  CD:  :CV  :  EV. 

But  as  E  approaches  C,  CV  and  EV,  approach  contin- 
ually to  a  ratio  of  equality,  therefore  also,  CE  and  CD 
approach  to  equality,  and  before  E  coincides  with  C, 
they  differ  from  each  other  less  than  by  any  assignable 
difference  ;  they  are  therefore  ultimately  equal.  But  Ev 
is  equal  to  DC,  therefore  CD,  CE,  and  Er,  are  all  ulti- 
mately equal. 

Q.  E,  D. 


PROPOSITION  I. 

The  curvatures  of  different  circles,  are    to  each  other 
inversely  as  their  diameters,  or  radii. 
Fig.  '^-      Let  DC  A,  touch  each  of  the  circles,  CEB,  CFG,  in  the 
point  C,  and  let  Ee,  F/,  be  parallel  to  CD. 
then,  sim,  triang.  Ce  :  CE:  :CE  :  CB, 
and         ''         "     C/:CF::CF  :  CG, 

(Ce=)  DE  .  CB=:CE2.  h  DF  .  CG=CF% 
DE  .  CB  :  DF  .  CG:  :CE-'  :  CF^. 
But,  if  DEF,  move  towards  CG,  CE  and  CF,  become 
ultimately  equal,  being  each  equal  to  CD, 
then  CE2=CF=^&  ..LE  .CB  =  DF.  CG  ; 

16.  therefore  DE:DF::CG:CB::rad. of  CFG:  rad.of  CEB; 
but  the  ratio  of  DE  to  DF,  is  the  ratio  of  the  curvatures  of 


6.  8. 


or  THE  CURVATURE  OF  CONIC  SECTIONS.  95 

the  circles,  CEB,  and  CFG  at  the  point  C  ;  (1.)  therefore 
their  curvatures  also,  are  as  CG  to  CB,  that  is,  they  are 
inversely  as  the  diameters  or  radii  of  the  circles. 

Q.  E.  D. 

CoR.  1. — The  curvature  of  the  same  circle  is  uniform, 
or  remains  the  same  in  every  part,  since  the  preceding  de- 
monstration is  applicable  to  any  point  of  the  circumfer- 
ence. 

Cor.  2. — Also  the  curvatures  of  equal  circles,  are  equal; 
since,  they  are  inversely  as  their  diameters,  which  are  equal. 

CoR.  3. — The  curvature  of  different  circles  may  dif- 
fer indefinitely,  and  may  approach  indefinitely  near  to 
equality.  Thus  if  B,  move  towards  G,  the  diameters, 
and  of  course  the  curvatures  of  the  two  circles,  approach 
continually  to  equality,  until  they  differ  from  each  other, 
less  than  by  any  assignable  quantity,  and  when  B  and  G 
coincide,  then  the  circles,  coincide,  and  the  curvatures  are 
the  same. 

ScHOL. — Hence,  the  curvature  of  circles  is  made  the 
measure  of  the  curvature  of  all  curves,  since  their  varying 
curvatures  are  easily  compared  with  the  uniform  and  defi- 
nite curvature  of  the  circle. 

Def.  2. — If  a  circle  touch  a  Conic  Section,*  so  that  no 
other  circle  can  be  drawn  between  it  and  the  curve,  it  is  cal- 
led the  Circle  of  Curvature  to  that  point  of  the  curve. 

Cor.. — If  there  can  be  one  Circle  of  Curvature,  to  any 
point  of  a  Conic  Section,  there  can  evidently  be  but  one. 
since  if  there  were  two,  one  would  pass  between  the  oth- 
er and  the  Section  — contrary  to  the  definition. 

ScHOL. — The  definition  given  by  Euclid  of  a  straight 
line  which  touches  a  circle,  is  not  applicable  to  a  circle 
which  touches  a  curve,  for  this  circle  may  cut  the  curve 
which  it  touches.  It  is  sufficient  that  they  both  touch  the 
same  straight  line,  in  the  same  point.     There  is  however, 

♦Circles,  and  curves  generally,  are  said  to  touch  each  other,  when  they 
^ouch  the  same  straight  line,  in  the  same  point. 


9«  OF  THE  CURVATURE  OF  CONIC  SECTIONS. 

a  striking  analogy  between  the  Circle  of  Curvature,  in  rela- 
tion to  the  curve  which  it  touches,  and  a  line  touching  a 
circle,  for  it  is  demonstrated  concerning  the  latter  (Euclid 
III.  i6  )  that  no  other  straight  line  can  be  drawn  between 
it  and  the  circle. 


PROPOSITION  II. 

If  from  the  vertex  of  a  Conic  Section  a  part  of  the  axis, 
be  taki  1)  equal  to  its  Parameter,  a  circle  described  on  this 
line,  will  be  the  Circle  of  Curvature  to  the  section  at  its 
vertex. 
^V  4.  That  is,  let  RAS,  be  a  Conic  Section,  AP  =  its  Param- 
eter, then  AFP  is  the  Circle  of  Curvature  to  the  curve  in  A. 

Fig.  4.       1.  If  CAGis  a  Parabola,  draw  AD  =  the  Parameter,  and 
DL  parallel    to  the  axis,  join   DP,  and  from  any  point   in 
the  diameter  of  the  circle,  draw  EFGIH,  perpendicular  to 
6  4     it  then  because  PA=AD,.'.PE  =  EI, 
then,  EF=  =  AE  .  EP= AE  .  EI. 
But  (I9.)EG==AE  .EH, 

the  efore  as  EI  is  less  than  EH,  EF^  is  less  than  EG^, 
and  EF  is  less  than  EG.     But  E  is  any  point  in  ihe  diam- 
eter AP.  therefore  every  part  of  the   circle   AFP  is  with- 
in the  curve  C'AG. 

2.  Let  any  line,  A^  be  taken  greater  than  the  Parameter, 
and  the  circle  AK  pcr^  be  described,  whose  centre  is  P,  the 
part  of  this  circle  K\g,  is  without  the  curve  CAG. 

Take  PO,  =the  Semi-Parameter,  and  draw  OK,  cut- 
ting the  curve  in  C,  and  the  circle  in  K,  from  C  draw  the 
tangent  CT,  and  also  AM  a  tangent  to  the  vertex,  join  PM, 
PC  is  perpendicular  to  CT,  (;Par.  VIII.  Cor.  4.) 

Also,  (Par.  VIII.  Cor.  2.)CM=MT,&  MT  is  less  than  MA, 

Therefore  CM  is  less  than  MA; 

But  PC  =  +CM=^=PM-^=MA2-HPA% 

And  as       CM  ^  is  less  than  MA-,  PA^  is  greater  than  PC% 
Therefore  PA=PK  is  greater  than  PC. 
That  is  the  circle  is  without  the  curve, 


OF  THE  CURVATURE  OF  CONIC  SECTIONS.  97 

If  Ap,  be  nearer  to  equality  with  AP,  or  if  p  be  suppo- 
sed to  move  towards  P,  O  will  approach  equally  to  A,  and 
OK  to  AM,  but  PK,  will  always  be  greater  than  PC,  un- 
til the  moment  that  p  and  P,  E  and  A,  and  the  two  circles 
coincide.  If  Ap  therefore  be  in  any  degree  greater  than 
the  Parameter,  the  circle  described  upon  it,  will  fall  with- 
out the  curve,  on  each  side  of  A,  the  point  of  contact  but 
if  it  be  equal  to  the  Parameter  it  will  fall  wholly  within  the 
curve, therefore  no  circle  can  bedraum  between  the  circle  AFP 
described  upon  the  Parameter,  and  the  curve  of  the  Parab- 
ola. In  other  words  it  is  the  Circle  of  Curvature,     (def.  2.) 

q.  E.  D. 


ScHOL. — The  same  may  be  demonstrated,  in  nearly 
the  same  manner,  concerning  the  Ellipse  and  Hyperbola, 
if  DL,  instead  of  being  drawn  parallel  to  the  axis,  be  drawn  Fig. 
to  its  other  vertex  as  DB.  (Hamilton's  Conic  Sections, 
B.V.  Prop.  XVI,  and  XVII.)  The  demonstrations  are 
omitted  here,  as  the  properties  will  be  included,  in  some 
more  general  propositions  which  follow. 

It  is  evident  from  the  preceding  demonstration,  that  ajiy 
circle,  ^rea^er  than  the  Circle  of  Curvature,  is  without  the 
curve  of  the  section,  on  each  side  of  the  vertex.  And 
yet  such  a  circle  may  approach  the  Circle  of  Curvarture,  in- 
definitely; it  may  differ  from  it  less  than  by  any  as- 
signable difference,  still  the  Parabola  is  within  the 
circle,  until  the  moment  that  the  latter  coincides  with  its 
Circle  of  Curvature.  It  is  for  this  reason,  that  the  curve 
is  said  to  coincide  with  its  Circle  of  CurvalurCj  at  the  point 
of  contact. 

CoR. — The  curve  and  its  Circle  of  Curvature  have  the 
same  currature,  at  the  point  of  contact. 

For  the  curve  has  not  a  greater  curvature,  than  its  Circle 
of  Curvature,  because  the  latter  is  wholly  within  it.  It 
has  not  a  less  curvature,  because  then  a  circle  might  be 
described  between  it  and  the  curve.  (I  Cor.  3.)  Since 
then,  its  curvature  cannot  be  either  greater  or  less  than 
that  of  its  Circle  of  Curvature,  it  must  be  the  same,  and 
their  evanescent  subtenses  which  measure  their  curvatures 
must  be  equal. 


OF  THE  CURVATURE  OF  CONIC  SECTIONS. 

ScHOL. — This  proposition  and  corollary,  compared 
with  Prop.  I,  illustrate  the  nature  of  Curvature,  and  its  dif- 
feren  degrees.  In  two  unequal  circles,  whose  curvatures 
consequently  are  unequal,  DE  and  DF,wliich  are  the  meas- 
ures of  these  curvatures,  do  not  approach  to  a  ratio  of 
equality,  (when  DF  moves  to  CG,  but  remain  ultimately 
t/ne^'wa/,  in  the  definite  ratio,  of  CG  to  CB;  consequen  ly 
the  angles  DCE,  DCF,  of  which  DE  and  DF  are  the  sub- 
tenses, and  to  which  they  are  nearly  proportional,  are  ulti* 
raately  unequal, in  the  same  ratio.  The  curves  EC,  and  FC, 
therefore  touch  the  straight  line  AD.  differently;  since  they 
approach  the  point  of  contact,  with  different  inclinations. 

But  DE,  and  DF,  the  subtenses  of  the  curve,  and  its  Cir- 
cle of  Curvature  do  approach  to  a  ratio  of  equality  for  the 
curves  have  both  the  same  curvature,  and  DE  and  DF 
are  therefore  ultimately  equal,  that  is,  the  corres^ponding 
point,  in  the  curve  and  its  t  ircle  of  Curvature,  are  ultimate- 
ly at  the  same  distance  from  the  tangent,  and  tlie  curves  EG 
and  FC,  approach  to  perfect  coincidence  as  the  points 
F  approach  the  point  of  contact;  they  meet  the  tangent 
with  the  same  inclination,  and  have    the  same  curvature. 


Def.  4. — The  Radius  of  Curvature,  the  Diameter  of 
Curvature,  and  the  Chord  of  Curvature,  of  any  curve,  are 
the  ladius,  the  diameter,  and  the  chord  of  the  Circle  of 
Curvature,  which  belongs  to  the  curve  in  that  point. 


PROPOSITION  III. 

In  the  parabola,  the  Chord  of  Curvature  which paS' 
ses  through  the   Focus,  is   equal  to  the  parameter  of  the 
diameter  which  passes  through  the  point  of  contact. 
Fig.  5.      ThatisCV=:4CF=P. 

Take  any  point  in  the  curve,  as  E,  draw  ED  parallel  to 
CV,  and  Ef,  parallel  to  CT,  cutting  CV,  in  x,  forming  the 
parallelogram,  Dx;  the  triangle  Cxv,  being  similar  to  CFT, 
(because  xv  is  parallel  to  CT,  and  Cv  to  TF)  and  FT  be- 
ing equal  to  FC,  (Par.  VII,  Cov.)  Cx=Cv,.\Cv=ET). 
Now  if  D  approaches  to  C,  indefinitely,  x  and  v  approach 
to  C  also,  and  then  Ev  approaches  indefinitely  to  equality 
with  EC.     Hence  ultimately, 'Ev=EC. 


OF  THE  CURVATURE  OF  CONIC  SECTIONS.  99 


6.  8. 


Then,  (in  the  cir  as  Fig.  1.)  ED  :  EC :  :EC  :  CV, 

And   (bee.  Cv=ED)  Cv  :  Ev.'.Ev  :  CV, 

Th-'efore,  Cv     CV=Ev^ 

Butn9.)  Cv.P'-rEvS 

.-.  (Par.  XIV.)  P'=CV,  and  P'=4FC. 

q.  E.  D. 


PROPOSITION  IV. 

A  parallelopiped  whose  base  is  the  square  of  the  Diam- 
tier  of  Curvatvre^  and  its  height  is  the  distance  of  the  Fo- 
cus from  the  vertex  of  the  Parabola,  is  equal  to  four  times 
the  cube  of  the  distance  of  the  Focus  from  the  point  of  con- 
tact. 

That  is  CB^  .FA=4FC3. 

Let  CB  be  drawn  perpendicular  to  to  the  tanget  CT,  it 
is  the  Diameter  of  the  Circle  of  Curvature,  CVB.     Also 
FG,  perpendicular  to  CT,  is  parallel  to  CB. 
Hence,  sim.  tri.        CB  :  CV : :  CF  :  FG, 
And  CB^  :CV-::CF2  :  FG^ 

But  FG2  =  FC.FA, 

Therefore   CB=  :  CV^::CF"^  :  CF  .  FA::CF  :  FA, 
And  CB2  .FA  =  CV=^.CF=4CF2  .CF=4CF^ 

q.  E.  D, 

CoR, — As  FA,  the  height  of  the  parallelopiped,  is  a 
constant  quantity,  the  parallelopiped  varies  as  its  base  :  that 
is,  as  (..B-,  which  therefore  varies  as  4FC'.  In  other  Sup. 
words,  the  square  of  the  Diameter  of  curvature^  varies  as 
the  cube  of  the  distance  of  the  Focus  from  the  point  of  con- 
tact."^ 

As  this  point  therefore  becomes  more  distant,  the  Di- 
ameter, and  Radius  of  Curvature  increase,  and  the  curva- 
ture itself  diminishes.   (See  also  Parabola,  Prop.  I.  Schol.) 


3.  1.9- 


3.  8. 


*  Algebraically  (Par.  VIII.  Cor.  5.)  FG2=FA  .  FC,  and  since  FA  con- 
stant. 

••.  FGa  varies  as  FC  .-.  FCsvariesas  FGs. 

But  C  B  2  varies  as  FC  2  varies  as  FC  3  or  as  FG  5  ; 

.-.  CB  varies  as  FG3,  or  CX3. 

That  is,  the  Diameter,  or  Radius  of  Curvature,  varies  as  the  cube  of  the 
normal. 


1^  OF  THE  CURVATURE  OF  CONIC  SECTIOxNS. 

ScHOL. — When  the  point  of  contact  is  the  vertex  of  the 
Parabola,  the  diameter  of  Curvature  coincides  with  the 
chord  which  passes  through  the  Focus,  and  is  equal  to  it. 
In  that  case  therefore  the  Diameter  of  Curvature  is  equal 
to  the  Parameter  of  the  axis,  as  was  also  demonstrated  in 
Prop.  11. 


PROPOSITION  V. 

In  the  Ellipse,  the  Chord  of  Curvature,  which  passet 
through  the  centre,  is  equal  to  the  parameter  of  the  diam- 
eter which  passes  through  the  point  of  of  contact. 

That  is  CL=P',  or  CG  :  HK  :  CL  are  continued  pro- 
portionals. 

Fig.  6.        For  ED  and  Er,  being  drawn  parallel  to  CG  and  CD, 
then  (Ell.  XXI.)   Cv  ,  vG -.  Er^ :  iCG^  :  HK-\ 

Now  as  E  approaches  C,  Er  approaches  to  equality 
with  EC,  and  dG  to  equality  with  CG,  as  their  limiting  ra.- 
tio.     They  are  therelore  ultimately  equal. 

Then,  by  substitution,    ED  .  CG  :  EC^ ;  iCG^HKS 
but  the  chord  EC-'=ED  .  CL, 

therefore  ED  .  CG  :  ED  .  CL:  iCG^  :  HK^ 

and  CG:  CLiiCG^-.HKS 

CL:  CG::HK=:CGs 

HK.CL:HK.CG::HK=:  CG% 
HK.CL:HK2::HK.  CG  :  CG% 
.   ,     therefore  CL  :  HK:  :HK  :  CG, 

CL  .  CG=HK-^;  but  P  .  CG=HK% 
CL=F:    and  CG  :  HK  :  CL,  are  continued 
proportionals. 

Q.  £.  />. 


OF  THE  CURVATURE  OF  CONIC  SECTIONS.  iQl 


PROPOSITION  VI. 

The  diameter  of  curvature  (CO)  is  a  third  proportional 
to  twice  CP  and  HK.  That  is  2CP  :  HK  :  CO,  are  con- 
tinued proportionals,  or  2CP  .  CO=HK2. 

For,  sim.  tri.         CO  :  CL : :  Cc  :  OP : :  2Cc  :  2CP, 

therefore  (V)        CO  .  2CP=(2Cc)  or  CG  .  CL=HK% 

Q.  E.  D. 


PROPOSITION  VII. 

The  chord  of  curvature  which  passes  through  the  Focus, 
is  a  third  proportional  to  the  transverse  axis,  and  the  diam- 
eter conjugate  to  that  vi-hich  passes  through  the  point  of 
contact 

That  is     AB  :  HK  :  CV,    are  continued  proportionals. 

For,  Sim.  tri.  CV  :  CO:  :CP  :  CN:  :2CP  :  2CN, 
But     (XI)      CN=Ac,  and  2CN=AB, 

CV:C0::2CP:  AB, 
and  (VI)  CV  .  AB=CO  .  2CP  =  HK-% 

or  AB:HK::  HK:CV. 

Q.  E.  D. 

CoR.  1. — When  the  point  of  contact,  is  the  vertex,  the 
diameter  of  curvature,  the  chord  which  passes  through  the 
centre,  and  the  chord  which  passes  through  the  Focus,  all 
coincide  with  each  other,  since  they  all  coincide  with  the 
axis  ;  they  are  therefore  equal,  and  the  diameter  of  Curva- 
ture is  therefore  a  third  proportional  to  the  Transverse  and 
Conjugate  axes,  that  is,  it  is  equal  to  the  Parameter.  (See 
Prop.  II.  Schol.) 

Cor.  2. — When  the  point  of  contact  is  the  extremity  of 
the  Conjugate  axis,  the  Transverse  axis  is  then  conju- 
gate  to  that  which  passes  through  the  point  of  contact ;  and 
then  the  equation  becomes 

CV  .  AB  =  ABS  and  CV=AB. 

That  is,  the  chord  of  curvature  passing  through  the  Fo- 
cus, then  equals  the  Transverse  axis. 

14 


10S  OF  THE  CURVATURE  OF  CONIC  SECTIONS. 

CoR>  3. — The  chord  of  curvature  which  passes  through 
the  centre,  is  to  that  which  passes  through  the  Focus,  as 
the  Transverse  axis,  is  to  the  diameter  which  passes 
through  the  point  of  coi^tact. 

For  (demonstration)     C  V  :  CO : :  2CP  :  AB, 
and  (VI)  CL:  CO::2CP:CG, 

CL:  CV::AB:CG. 

Cor.  4. — The  diameter  of  curvature,  the  chord  which 
passes  through  the  centre,  and  the  chord  which  passes 
through  the  Focus,  are  inversely  as  the  parts  cut  off  from 
each,  by  the  diameter  conjugate  to  that  which  passes 
through  the  point  of  contact. 

For  (Cor.  2)     CL  :  CV:  :AB  :  CG:  :CN  :  Cc, 
and  (VI)  CO  :  CL:  :CG  :  2CP:  :Cc  :  CP, 

5.20.  .:.  (  CO  :  CL  :  CV,  are  as 

I CN  :  Cc  :  CP. 

CoR.  5. — The  diameter  of  Curvature,  varies  as  the 
cube  of  the  diameter,  conjugate  to  that  which  passes 
through  the  point  of  contact. 

For  (Ell.XVl)  2CP  .  HK=AB  .  ab, 
but     (VI)  2CP  .  CO=HK», 

2CP  .  HK  :  2CP  .  CO  : :  AB  .  a6  :  HK% 
or  HK  ;  CO  : :  AB  .  ah  :  HK^,  '     . 

.-.  the  parallelopiped  CO  .  AB  .  ab=EK\ 
But  the  base  AB  .  ab  is  constant, 
CO  varies  as  HK^. 

Consequently  the  curvature  itself  (II)  at  any  point  is  in- 
versely  as  the  diameter  conjugate  to  that  which  passes 
through  that  point,  it  is  iherefore  least  at  the  extremity  of 
the  Conjugate  axis,  and  greatest  at  that  of  the  Transverse. 

Cor.  6. — The  diameter  of  Curvature,  at  the  intermedi- 
ate points,  varies  as  the  cube  of  the  normal,  CI. 

For  (Ell.  XV)  CI  .  CP=Ca=»  .-.  CP  varies  as  CI, 
and  (Ell.  XVI)  CP  .  HC=Ac  .  ca  .'.  CP  varies  as  HC, 

CI  varies  as  HC,  and  CP  asHC,  asHK% 
.;.     CO  which  varies  (Cor.  5)  as  HK^,  varies  as  CP. 


OF  THE  CURVATURE  OF  CONIC  SECTIONS. 


PROPOSITION  VII L 

In  the  hyperbola,  as  in  the  Ellipse,  the  chord  of  cur- 
vature, which  passes  through  the  centre,  is  equal  to  the 
parameter  of  the  diameter  which  passes  through  the  point 
of  contact. 

That  is      CG  :  HK  :  CL,  are  continued  proport's. 

For         Ci  .tG  :Er«::CG2  :  HK% 
whichisulti.  ED  .  CG  :  ED  .  CL::CG»  :  HK^ 
and        CG  :  CL::CG2  :  HK^ 
CL:CG::HK2  :  CG% 
HK.CL  :  HK.CG::HK»  :  CG% 
HK  .  CL  :  HK» : :  HK  .  CG  :  CG% 
CL  :  HK::HK  :CG. 
But  (19)  P'  :  HK::HK  :  CG. 

.-.      P'=CL,  and  CG  :  HK  :  CL,  contin'd.  prop'Is. 

Q.  E.  D, 


PROPOSITION  IX. 

The  Diameter  of  Curvature,  is  a  third  proportional  to 
2CP  and  HK ;   or  2CP  :  HK  :  Co,    are  continued  prop's. 

Forsim.  tri.  CO:  CL:  :Cc  :  CP:  :2Cc  :  2CP, 
CO  .  2CP=CL  .  CG=HK% 

Q.  E.  D, 


PROPOSITION  X. 

The  chord  of  curvature  which  passes  through  the  Focus, 
is  a  third  proportional  to  the  transverse-axis,  and  the 
diameter  conjugate  to  that  which  passes  through  the  point 
of  contact. 

That  is    AB  :  HK  :  CV,  are  continued  proportionals, 

Insim.  tri.  CV  :  COl.CP  :  CN::2CP  :  2CN==AB 
.'.       CV.AB=C0.2CP=HK^ 
i»r       AB  :  HK  :  CV,  are  coatinued  proportionals. 

Q.  E.  D, 


103 


104  ^^  THE  CURVATURE  OF  CONIC  SECTIONS. 

Cor.  1. — At  the  vertex  of  the  Hyperbolas,  the  diame- 
ter of  curvature,  coincides  with  the  chord  which  passes 
through  the  centre,  and  the  chord  which  passes  through 
the  Focus,  since  they  all  coincide  with  the  axis;  it  is  there- 
fore equal  to  the  Parameter.     (See  Schol.  Prop.  II.) 

The  diameter  of  curvature  varies  as  the  cube  of  the 
normal  (CI),  and  therefore  the  curvature  itself  is  inversely 
asCP. 

Schol.— The  analogy  between  the  curvature  of  the  Ellipse 
and  Hyperbola  is  complete,  since  the  corresponding  prop- 
ositions are  capable  of  being  expressed  in  the  same  words. 
The  terms,  however,  as  was  remarked  in  reference  to  the 
general  properties  of  the  two  sections,  have  somewhat  dif- 
ferent significations  ;  and  in  particular  the  normal^  the 
cube  of  which  is  proportional  to  the  radius  of  curvature,  in- 
creases more  rapidly  in  the  Hyperbola  than  in  the  Ellipse, 
and  never  decreases  again,  after  reaching  a  maximum^  as 
it  does  in  the  Ellipse.  The  curvatures  of  the  two  sections 
therefore  are  widely  different.  That  of  the  Hyperbola,  con- 
stantly decreases,  and  the  curve  never  returns  into  itself. 
When  the  two  conjugate  axes,  are  unequal,  the  curvature  is 
greatest  at  the  vertex  of  the  greater  axis,  as  in  the  Ellipse, 
and  if  the  two  axes  of  a  Hyperbola  are  equal  to  the  two 
axes  of  an  Ellipse,  each  to  each,  the  curvature  at  their 
vertices,  in  one  section  is  the  same  as  in  the  other. 

If  the  two  axes  of  a  Hyperbola  are  equal  to  each  other, 
(that  is  if  the  Hyperbolas  are  Equilateral,)  in  which  case 
the  section  corresponds  to  the  circle  into  which  the  El- 
lipse passes  as  its  two  axes  become  equal,  the  curvature 
at  the  fourvertices  of  the  Conjugate  Equilateral  Hyperbolas 
is  the  same,  and  is  equal  to  that  of  a  circle,  whose  diame- 
ter is  equal  to  the  axis  ;  and  the  diameter  of  curvature  to 
any  other  point,  is  as  the  distance  of  that  point  from  the 
centre.  The  curvature  itself  therefore  of  Equilateral  Hy- 
perbolas is  the  same,  in  any  point,  as  that  of  a  circle, 
whose  circumference  is  at  an  equal  distance  from  its  cen- 
tre. In  other  words,  in  both  figures,  the  curvature  is  al- 
ways inversely  as  the  distance  of  the  curve  at  that  point, 
from  the  centre.  In  this  respect,  therefore,  there  is  a 
striking  analogy  between  an  Equilateral  Hyperbola,  and 
the  circle,  which  may  be  considered  as  an  Equilateral  El- 
lipse,    The  difference  however,  between  the  two  curves 


OF  THE  CURVATURE  OF  CONIC  SECTIONS.  105 

is  great,  arising  from  the  fact,  that  in  one  the  curve  is 
convex  towards  the  centre,  and  in  the  other  it  is  con- 
cave. 

In  a  general  connparison  of  the  curvature  of  the  circle, 
with  that  of  the  other  Conic  Sections,  the  distinguishing 
peculiarity  of  the  former  is,  that  its  curvature  is  uniform^ 
remaining  the  same  in  every  point  of  its  circumference, 
while  that  of  the  three  other  sections  changes  continually, 
and  is  never  the  same  at  two  successive  points.  In  all  of 
them,  the  curvature  is  greatest  at  the  vertex  of  the  sec- 
tions, and  decreases  according  to  a  given  law,  as  any  point 
is  removed  from  the  vertex. 

This  continual  Variation  of  Curvature ^  as  it  is  termed, 
is  the  reason  why  no  definite  portion  of  any  Conic  Sec- 
tion, coincides  with  its  Circle  of  Curvature.  If  the  curva- 
ture of  the  section,  remained  the  same,  orwas  uniform  for 
any  distance,  that  portion  would  be  a  part  of  a  circle,  and 
would  consequently  coincide  with  its  Circle  of  Curvature, 
since  it  has  been  demonstrated  that  the  latter  has  the  sanip. 
curvature  as  the  curve  of  the  section  at  the  point  of  con- 
tact. Since  however  the  curvature  of  a  Conic  Section, 
as  has  been  shown,  never  continues  the  same,  in  two  suc- 
cessive points,  but  varies  continually,  the  curve  immedi- 
ately deviates  from  the  tangent  more  or  less  than  the  Cir- 
cle of  Curvature  does,  and  therefore  passes  in  the  former 
case  within,  and  in  the  latter  without  the  circle.  As  the  cur- 
vature in  all  the  sections,  is  greater  at  the  vertex,  than  in 
any  other  point,  the  whole  section  passes  without  the  Circle 
of  Curvature  to  that  point ;  fora  similar  reason,  the  Ellipse  is 
wholly  within  the  Circle  of  Curvature,  to  the  curve  at  the  ex- 
tremity of  its  Conjugate  axis:  and  in  all  the  sections,  that  fj^  7. 
part  of  the  curve  which  is  towards  the  vertex  from  any  point, 
being  more  curved  than  at  the  point,  passes  within  the  Cir- 
cle of  Curvature  to  that  point,  while  the  part  more  remote, 
being  less  curved,  passes  without  it. 


APPENDIX  TO  CONIC  SECTIONS. 

J.  Similar  and  Sub-contrary  Sections. 


Def.  1. — If  a  point  be  taken  above  the  plane  of  a  cir^ 
clet  and  one  extremity  of  a  straight  line,  remain  in  that 
point,  while  the  line  moves  around  in  the  circumference 
of  the  circle,    the   figure  described  is  called   an   Oblique 

PI.  VII.  ^onc. 

Fi?.  8.  as  VLK. 

2. — If  either  of  the  Conic  Sections  be  supposed  to  re- 
volve around  its  axis,  which  remains  fixed,  the  solid  thus 
generated,  may  be  called  a  Conoid,* 

Conoids  receive  specific  names,  from  the  diflferent  fig- 
ures by  which  they  are  described  ;  as  Paraboloids,  Ellip- 
soids and  Hvperboloids. 
f^'^;^^!'  '     As  GAH,  AHBG,  and  GAHO. 


PROPOSITION  I. 

All  the  sections  of  a  cone,  nvide  by  parallel  planes,  are 
similar  figures. 


tlZ. 


Parabolic  sections,  AD  :  ad:  :DG  :  d^ 
Fig. 2.       Ellipses,  AB  :  ab:: A-B'  :  a'b'. 


,  *  Perhaps  this  use  of  the  term  is  not  authorized  :  but  some  general  name 
was  needed  in  the  following  propositions,  and  it  seemed  most  easy  and 
proper  to  extend  the  application  of  a  term  already  in  use.  Archimedes 
gives  the  name  Conoids  to  Paraboloids  and  Hyperholoids,  (probably  be- 
cause they  obscurely  resemble  cones,)  and  for  a  similar  reasofn  to  Elif^ 
soids,  he  o;ives  the  name  of  Spheroids. 


APPENDIX  TO  CONiC  SECTIONS.  lOT 

Hyperbolic  Sections,  AB  :  conj.  of  AB y.ab  :  conj.  of  Fig.  3. 


For,  the  conj.of  AB=a  mean  proportional  between  AQ 

and  BR, 
and  the  conjugate  of  ff6=a  mean  proportional  between  ag 

and  hr. 
therefore     AB  :  its  conjugate  Wnb  \  its  conjugate. 

Cor.  1. — If  a  plane  perpendicular  to  thr  triangular 
section,  pass  through  the  vertex  of  the  cone,  cutting  the 
base,  it  will  cut  off  similar  segments,  from  similar  parallel 
sections  ; 

that  is,  in  the  Parabola,  AD  :  DG:  \ad  :  dg,  F,g   ,^ 


J5         5» 


Ellipse,     AD  \T>E:\Mdxde.  "  ?. 

and  „  BD:  DE::B'^:  (/e. 

AD:  A'(/::AB:A'B'. 

DE  :  de'.'.ab  :  ah', 

.,  Hyperbola     AD  \  DGWad  :  dg,  ?<    3 

BD:DG::W:rf^, 

AD  :  a(/::AB  :  «^, 

DG  :  dg:  : conjugate  of  y^B  !  conjugate  o(  ab. 

Cor.   2. — Similar  segments  may  be  cut  from  all  Parab- 
olas, that  is,  all  Parabolas  are  similar  sections.  y^i'% 
For,  if  there  be  taken  DG-^  :  DG'^^AD  :  AH,  '  ^^ 
then                         AD  :DG::AH:HK. 

Cor.  3. — Similar  Polygons  may  be  inscribed  in   similar 
segments  of  similar  sections. 


PROPOSITION  H. 

If  any  Conoid,  be  cut  by  a  plane  parallel  to  its  base,  01' 
to  the  plane  of  its  revolving  axis,  the  section  will  be  a  cir 
ofle.  Fig.  4.6. 

Thus  MZN  and  HS^^G  are  circular  sections.  and  6. 


]Qg  APPENDIX  TO  CO?JlC  SEGT10>'3. 


PROPOSITION  III. 

If  any  Conoid,  be  cut  by  a  plane,  coinciding  with  its  ax- 
is, the  section  will  be  a  figure,  equal  and  similar  to  that,  by 
the  revolution  of  which  the  Conoid  was  generated. 
Kig.  4.5.     For  it  coincides  with  the  revolving  figure,  in  one  of  its 
positions.  . 

Thus  GAHO,  may  be  considered  a  section  of  the  Co- 
noids. 


PROPOSITION  IV. 

If  a  Conoid  be  cut  by  a  plane  parallel  to  its  fixed  axis, 
the    section   will  be  of  the  same  kind  with   that   by    the 
revolution  of  which,  the  Conoid  was  generated,  and  simi- 
I  a?'  to  it. 
^'•g-  4       Suppose  a  plane  GAHO  to  coincide  with  the  axis  produ- 
^•*"     "  cing  the   section  by  which  the  conoid  was  described,  and 
cutting  the  given    plane  at  right  angles.     Let    there    be 
two  circular  sections  MZN  and  HS^G   perpendicular  to 
the  fixed  axis,  and  therefore  at  right  angles  to  each  of  the 
two   other  planes.      The    mutual     intersection    of  these 
planes  RZ  and  DG  will  be  ordinates,  both  in  the  circular 
sections,  and  in  the  first  sections,  and  it  will  be   easy   to 
prove, 
fi?-  4.       1.  In  the   Parabola,   that  the  squares  of  the  ordinates 
are  as  their  abscisses  (See  Par.  Prop.  1.) 
That  is,  aR  :  ad.'.KZ''  :  dg^. 

And  that  the  section,  is  similar  to  that  which  coincides 
with  the  axis,  or  to  that  by  which  the  Conoid  was  genera- 
ted. 

That  is,  ad  :  dg:  .AD  :  DG. 

Fig.  5.       3.  In  the  Ellipse,  that  the  squares  of  the  ordinates  are 
as  the  rectangles  of  theirabscisses  (Ellipse  Prop.  I.) 
That  is    aR.RB:ad.db::RZ^:dg'. 

and  that  AD  :  DG '.'.ad  :  dg, 
and     AB  :ab::HG  iSldg. 


APPENDIX  TO  CONIC  SECTIONS  1(J9 

u  ill  the  H}perbola,  if  a  plane  parallel  to  that  in  which 
the  axes  of  four  Conjugate  Hyperboloids  are  placed,  cat 
them,  it  may,  in  a  similar  manner,  be  proved,  that  the  two 
opposite  sections,  are  Hyperbolas,  and  in  every  respect 
equal  and  similar  to  each  other,  and  that  the  four  sections 
are  Conjugate  Hyperbolas,  similar  to  the  four  by  whicli  the 
Hyperboloids  were  generated. 

That  is  cR'  -ca^  :  cd'  -ca^  :  .'RZ^  :  dg',  ''.f  ^^-^ 

and         AB  :  abr.A'B'  la'b'. 

Cor, —  If  a  plane  cut  two  opposite  Hyperboloids,  be- 
ing perpendicular  to  the  plane  in  which  their  axes  are  pla- 
ced, and  pasjifii;  through  their  centre  (and  therefore  coin- 
ciding with  a  diameter  of  the  Hyperbolas,  by  which  the 
Hyperboloids  were  generated,^  the  sections  will  be  two 
opposite  H\perbo|as  equal  to  each  other  in  every  respect, 
but  not  similar  to  the  generating  Hyperbolas. 

Thati^OIQ  is  a  Hyperbola,  orim  .  mH  :  I5  .  sH  iini^  :  sv", 
and  the  opposite  section  will  be  equal  and  similar. 

If  then  another  plane  cut  the  two  Conjugate  Hyperbo- 
loids in  the  same  manner,  but  coinciding  with  KB"  the  di- 
ameter which  is  conjugate  to  the  other,  this  section  will  al- 
so form  two  opposite  Hyperbolas  which  will  be  conjugate  to 
the  other  two. 

and  III-  :  KB"-  :  'Am  .  mH  :  mi-.  Fig-  7. 


PROPOSITION  V. 

if  in  any  Conoid,  a  plane  pass  so  as  to  intersect  all  sides 
of  it,  the  section  in  every  case,  will  be  an  Ellipse.  p. 

In  addition  to  the  pianos  supposed  in  the  last  figure,  lets,  aidi, 
a  plaiie  coincide  with  RZ  (the.  mutual  intersection  of  the 
circular  section  MZN,  and  the  section  parallel  to  the  axis, 
ttZgd)  and  let  it  cut  the  axis  of  the  Conoid  and  all  its 
sides  ;  it  can  easily  be  shewn  that  this  oblique  section  has 
in  every  case  the  properties  of  an  Ellipse,  (See  Ellipse, 
Prop.  1.) 

That  is  XR  .  RO  :  XP  .  PO:  :RZ^  :  PS^. 

CoR. — All   sections    parallel    to   any    oblique   section 
are  Ellipses  an^j  are  similar. 

15 


no  APPENDIX  TO  CONIC  SECTIONS 


PROPOSITION  VI. 


Every  section  of  an  oblique  cone,  which  is  parallel  to 
its  base,  is  a  circle.  Every  section  which  is  sub-contrary 
to  the  base  is  a  circle. 

Every  other  section  is  an  Ellipse,  and  all  the  Elliptic 
sections  which  2iVQ  between  the  sections  parallel  and  sub- 
contrary  to  the  base,  may  be  called  Ellipses  sub-contrary 
to  all  other  Elliptic  sections  of  the  same  cone,  that  is  their 
Transverse  and  Conjugate  axis,  are  placed  in  contrary  po- 
sitions. 

1.  Sections  parallel  to  the  base,  are  proved  to  be  cir- 
cles, by  shewing  that  all  lines  drawn  in  them,  from  the  ax- 
is of  the  cone  to  its  circumference  are  equal. 
Fig.  8.  that  is  cB=BA. 

2  If  a  plane,  cut  the  cone  so  as  to  make  the  same  an- 
gles with  the  axis  and  with  the  opposite  sides  of  the  cone, 
that  are  made  by  a  section  parallel  to  its  base,  but  in  a 
contrary  order,  it  is  called  a  sub-contrary  section,  and  is 
proved  to  be  a  circle,  as  in  case  first. 

That  is,  if  VBA=VAB  and  cd  be  the  mutual  intersec- 
tion of  the  circular  section,  parallel  to  the  base  B'cR', 
and  the  given  sub-contrary  section,  BA,  then  Be  .  c\=cd^. 

3.  That  all  other  sections  are  Ellipses,  is  proved  as  by 
Prop.  I  Ellipse. 

4.  Of  any  section  between  the  parallel  and  sub-contrary 
sections  as  AB',  AB'  is  the  conjugate,  or  smaller  axis. 

',  ^  ScHOL. — Those  circular  sections  which  are  parallel  to 
the  base,  and  those  which  are  sub-contrary,  in  an  oblique 
cone,  are  limits  between  sub-contrary  Elliptic  sections, 
and  hold  the  place  of  transition  from  one  of  those  Elliptic 
sections  to  the  other. 

Thus,  the  sections  AB,  AB  are  limits  hetvfeen  AB'  and 
AB' ,  two  Elliptic  sections,  one  on  each  side,  but  having 
their  axes  in  a  contrary  position. 


APPENDIX  TO  CONIC  SECTIONS  Hi 


PROPOSITION  VII. 

Every  section  of  a  Cylinder,  parallel  to  its  base  is  a  cir- 
cle, every  other  section  is  an  Ellipse. 

The  first  case  is  evident. — The  second  is  proved  in  a 
manner  similar  to  Prop.  I,  of  the  Ellipse.  Eig 

That  is,  BD  .  DA  :  BS  .  SA:  iDE^'  :  SZ«. 

ScHOL. — The  demonstrations  of  the  preceding  proposi- 
tions have  not  been  given  in  full,  on  the  supposition  that 
to  persons  well  acquainted  with  the  propositions  on  Con- 
ic Sections,  the  truth  of  them  will  be  almost  immediately 
evident ;  to  others,  it  may  afford  a  useful  and  agreeable  ex- 
ercise, to  make  out  the  demonstrations,  by  deducing  them 
from  the  properties  of  the  curves  previously  demonstrated. 


APPENDIX  TO  CONIC  SECTIONS, 


n.    COMPARISON    OF    CONIC    SECTIONS, 


^i$§i 


PROPOSITION  I. 

•  A  Parabola  is  equal  to  four  thirds  of  a  Triangle,  having 

the  same  base  and  altitude.     Or  it  is  two  thirds  of  a  circum 
PI  VIII  scribmg  parallelogrann. 
Fig.  10.'      That  is  Par  BAC=|  Tri.  BAG. 

Let  the  base  of  the  Parabola  BAG,  be  bisected  contin- 
ually, dividing  it  into  the  equal  parts,  BR,  RG,  GS,  SD, 
DX,  XE,  EY,  YC,  and  let  lines  be  drawn  through  the 
intersections  parallel  to  the  axis  AD,  meeting  the  curve. 
Let  GA  be  drawn  parallel  to  the  axis,  and  made  of  such 
length:  that  Ca  :  DA::4  :  3. 

Connect  aB  also  AB,  AH,  HB,  AG,  AF,  FG,  form- 
ing the  inscribed  polygon  BHAFG.  Then  RP  pass- 
ing throng]^  that  division  of  the  base,  which  is  next  to  B, 
and  meeting  Ba  in  d  and  BH,  in  / ;  and  GH,  meeting  oQ^ 
AB,  in  c  and  L ;  HQ  also  being  drawn  parallel  to  BG,  and 
Vq  parallel  to  BA,  and  DO,  HO  perpendicular  to  AB, 
Then  tri.  BaG:BAC::Gfl  :  DA::4  :  3, 
and  sim.tri.BaG=4B6D,  or  B6D  =  iBaG, 

BflG:6aGD::4:  3, 
therefore      6aGD=BAG. 

Again  tri.     BHA  :  BD A : :  HO'  :  DO : :  HL  :  AD, 

BHA-i-AFC  :  BAG:  :HL  (  =  AQ)  :  kD, 

::HQ»:BDs 
::  1      :  4. 

And         DGc6:  GD6a(=BAC)  ::    1       :    4. 
Therefore  BHA+AEC=DGcJ. 


APPENDIX  TO  CONIC  SECTIONS.  113 

In  like    manner  BPH-f-HMA  +  ANF+FNC  : 
BHA+AFC,  ::P/:  HL, 


P72  :  AL 


::RG=:  BG% 
::    1      :     4, 
BuiGRr^c  :  Gc6D(  =BH A  +  AFC) : !  1  :  4  ; 
Therefore  BPH  +  HMA  +  ANF+FVC  =  GRrfc. 

In  the  like  manner  if  the  parts  of  the  base  be  bisected 
continually,  and  the  sides  of  the  inscribed  polygon  multipli- 
ed in  the  same  proportion,  the  polygon  will  in  all  cases,  be 
equal  to  the  trapezium  cut  from  the  triangle  BaC,  by  the 
parallel  to  the  axis,  nearest  to  B.  But  the  polygon,  thus 
approaches  continually  and  indefinitely  towards  equality 
with  the  Parabola,  which  is  its  limit;  also  the  trapezium 
approaches  indefinitely  towards  the  triangle  BaC,  which  is 
evidently  its  limit;  and  since  the  polygon  is  always  equal  to 
the  trapezium,  the  lijnit  of  xhe  polygon,  that  is,  the  Parabo- 
la is  equal  to  the  limit  of  the  trapezium,  or  the  triangle 
BaC. 

If  this  conclusion  is  denied,  it  can  be  shown  that  the  de- 
nial leads  to  absurdity  ;  for  if  the  triangle  BaC.  for  instance 
were  supposed  to  be  greater  than  Parabola,  then  the  pro- 
cess could  be  carried  on,  until  the  trapezium  taken  from 
the  triangle,  should  be  also  greater  than  the  Parabok  ;  but 
the  trapezium  is  always  equal  to  the  polygon  inscribed  in 
the  Parabola,  and  therefore  always  less  than  the  Parabola. 

A  like  absurdity  follows  the  supposition  that  the  Parabo- 
la is  greater  than  the  triangle.  », 

Therefore  the  triangle  BaC.  is  equal  to  the  Parabola. 

But  the  triangle  BAC  =  5  BaC, 

.*.  also  the  inscribed  triangle  BAC  =  f  Parabola, 

Or  the  Parabola  =^  Triangle  BAC. 

Q.  E.  D. 

CoR.  1.  — Hence  if  two  Parabolas  have  equal  altitudes, 
they  are  to  each  other  as  their  bases. 

For  this  is  true  of  the  circumscribing  parallelograms,  or 
the  inscribed  trianglei.  ^  sf  15 

CoR.  2. — And  if  two  Parabolas  have  equal  bases,  thejr 
are  to  each  other  as  their  altitudes. 


6.  I. 


114  AFPEXDIX  TO  CONIC  SECTIONS. 

Cor.  3. — Any  two  Parabolas   are  to  each   other,  in  the 
ratio  compounded  of  their  bases  and  altitudes. 

Cor.  4. — Hence   similar  Parabolic    Sections    have  to 
each  other,  the  duplicate  ratio  of  their  bases,  or  altitudes. 


PROPOSITION  II. 

The  Ellipse  has  to  the  circle  which  is  described  upon 
Its  Transverse  Axis,  the  same  ratio,  which  the  Coujugate 
axis  has  to  the  Transverse  ;  — and  to  the  circle  described 
upon  its  Conjugate  axis,  the  same  ratio  which  the  Trans- 
verse has  to  the  Conjugate. 

PI.  VIII.     That  is  Ellip.   Aa  Bb  :  Cir.  AGE:  [ab  :  AB. 

Fig.  12.*     And  Ellip.  Aa  Bb  :  Cir.  agb:  :AB  :  ab. 

Let  AGB.  be  a  semi-circle,  described  upon  AB,  the 
Transverse  Axis,  of  the  Ellipse  AKB,  let  NHF,  MKG  be 
any  two  ordinates  in  the  Ellipse,  produced  until  they  meet 
the  circle,  join  GF  and  KH, 

Then  (Ell.  Ill  Cor.  2.)  MK:  MG:  iNH'-NF  ; 
Therefore  GF.  KH,  produced  will  meet  in  R,  in  MN 
produced.  Also,  il  GF  be  bisected  in  D,  and  the  ordinate 
DEI  be  drawn,  then  the  tangents  to  the  Circle  and  Ellipse 
at  the  points  D  and  E,  will  meet  CA  produced  in  the  same 
point  T.  (Ellip.  VI.  Conversely.) 
6.  1.        Then.  tri.  MGR  :  MKR : :  MG  :  MK, 

Also     "     NFR  :  NHR:  :NP  :  NH:  :MG  :  MK, 
5  ^9       .'.  trap  MGFN  :  MKHN:  :MG  :  MK. 

also  tri.     NEA:NHA::NF  :NH::MG:MK, 
and  "       MGB  :  MKB : :  MG  :  MK, 
.-.  PolAFDGB  :  aHEKB::MG:  MK::CA:Ca. 
6.  4.        But,  DT  is  parallel  to  OR, 

.-.  ER  is  parallel  to  KR  and  the  triangles  GDF,  KEH, 
are  each  greater  than  half  their  segments  GDF,  KEH, 
therefore  the  inscribed  polygons,  as  the  sides  are  multi- 
plied, will  approach  indefinitely  to  equality  with  the  semi- 
circle and  the  semi-ellipse,  as  their  limits. 

.-.Cir  :  Ellipse  : :  Pol.  AGB  :  Pol.  AKB:  :CA  :  Ca. 
PI  in       In  like  manner  Ellipse  A«B  :  Cir.  Dab:  :CA  :  Co. 
Fig.  2.  Q.  E.  D. 


APPENDIX  TO  COMC  SECTFOJ^S.  US 

Cor.  1. — Any  Ellipse  is  a  mean  proportional,  between 
ihe  circles  described  upon  its  two  axes. 

That  is,  Cir.  AGB  :  Ell.   AaBb : :  Ell.  AaBb  :  Cir.  agb. 

Cor.  2. — An  Ellipse  is  equal  to  a  circle  whose  diameter 
fs  a  mean  proportional  between  its  two  axes. 

For  let  Z  be  such  a  circle,  ^'=*  ^^ 

Then  cir.  AGB  :  Z :  :  AB  =^  :  A  B'  ^ 

Also,  cir.  AGB  :  Ellip.: :  AC  :  Ca, 

::AC-^  :  AC.Ca, 

::AB=  :2AC  .  2Ca=A'B  ^ 

CoR.  3. — Hence  any  circle  is  to  the  square  of  its  diam- 
eter, as  the  inscribed  Ellipse,  is  to  the  rectangle  of  two  axes. 

CoR.  4. — The  areas  of  any  two  Ellipses,   are  to  each, 
ether  as  the  rectangles  of  their  Transverse  and  Conjugate 
axes. 
That  is,Ellip.  A'aB'6  :  Ellip.  Aa'B6':  :A'B'  .  ab  :  AB  a'b',  Fi-.  14. 

ScHOL. — The  two  preceding  propositions  are  demon- 
strated after  the  manner  of  the  ancient  Geometers.  See 
Archimedes  "  De  Quadraturae  Parabolae"  and  ''De  Con- 
©idibus  et  Spheroidibus." 

It  is  called  the  method  of  Exhaustions,  and  is  much  ad- 
mired for  its  ingenuity  and  rigorous  exactness. 

"  Though  few  things,  says  Playfair,  more  ingenious  than 
this  method  have  been  devised,  and  though  nothing  could  be 
more  conclusive  than  the  demonstrations  resulting  from  it, 
yet  it  laboured  under  two  very  considerable  defects.  In 
the  first  place  the  process  by  which  the  demonstrations 
were  obtained  was  long  and  difficult,  and  in  the  second  place 
it  was  indirect,  giving  no  insight  into  the  principle  on  which 
the  investigation  was  founded." 

This  method  has  been  much  abridged  by  the  moderns, 
especially  by  Cavaleri,  in  the  method  of  Indivisibles,  a 
method  not  only  more  concise,  than  that  of  the  ancients, 
but  much  more  easily  and  extensivly  applicable  to  a  great 
variety  of  propositions,  and  at  the  same  time  no  less  rigo- 
rously exact,  when  properly  managed. 


f 


116  APPENDIX  TO  COiMC  SECTIONS. 


An  application  of  it  to  the  second  of  the  preceding  prop- 
ositions will  illustrate  its  nature,  and  prepare  the  student 
for  the  demonstrations  which  follow. 

f''?-  11-  Any  trapezium  as  KMGD,  is  to  the  corresponding 
trapezium  in  the  Ellipse  KLED,  as  the  sum  of  the 
parallel  sides  KM  and  DG,  is  to  the  sum  of  KL,  and  DE, 
for  each  trapezium  may  be  divided  into  two  triangles, 
which  will  be  to  each  other,  respectively  as  these  lines. 
Consequently,  all  the  trapeziums  taken  together,  or  the 
polygon  in  the  semi-circle,  will  be  to  the  polygon  in  the 
Ellipse,  as  the  sum  of  the  ordinates  in  the  circle,  is  to  the 
sum  of  the  ordinates  in  the  Ellipse,  that  is  (Euclid  V.  12,  as 
any  one  ordinate  in  the  circle,  to  the  corresponding  ordi  :;  te 
in  the  Ellipse;  in  other  words,  as  the  Transverse  to  the  Cc  n- 
jugate  axes.  But  by  increasing  the  number  of  or  .;jiates, 
and  consequently  the  number  of  sides  of  the  polygons,  the 
latter  approach  indefinitely  to  equality  with  the  circle  and 
Ellipse;  therefore  the  circle  is  to  the  Ellipse,  in  the  same 
ratio  as  the  polygon  in  the  circle,  to  the  polygon  in  the  El- 
lipse, that  is,  as  the  sum  of  the  ordinates  m  one,  to  the  sum 
of  the  corresponding  ordinate  in  the  other,  or  as  the  Trans- 
verse :o  the  Conjugate  axes. 

This  method  of  Indivisibles,  or  of  supposing  ordinates  to 
be  encreased  indefinitely  in  number  while  their  distances 
are  diminished  in  the  same  proportion,  often  appears  unsat- 
isfactory, not  only  on  account  of  the  abridged  mode  of  ex-' 
pression,  by  which  important  steps  in  the  demonstration  are 
omitted,  but  because  it  is  common  to  say,  that  the  spaces 
compared  are  made  up  of  an  infinite  number  of  parallel  or- 
dinates, whereas  it  is  evident  from  the  definitions  of  Geom- 
etry, that  no  possible  number  of  lines  can  "  make  up"  a 
space,  or  form  any  thing  but  a  line;  not  to  mention  the  in 
consistency  of  an  '*  infinite  number"  of  lines 

The  truth  is,  as  has  just  been  shown,  that  the  rectilineal 
spaces  compared,  have  to  each  other  the  same  ratio  which 
the  sums  of  corresponding  ordinates  have,  and  curvilineal 
spaces,  which  are  their  limits,  have  exactly  the  same  ratio. 
The  exact  truth  of  this  Proposition,  may  be  rigorously  de- 
monstrated in  all  cases  by  a  reductio  ad  absurdum. 


APPENDIX  TO  CONIC  SECTIONS.  ||7 

The  method  of  Indivisibles  then,  when  properly  consid- 
ered and  applied,  is  an  abridged  form  of  the  method  of 
Exhaustions;  the  ungeometrical  language,  so  often  used, 
and  before  alluded  to,  will  be  avoided  in  the  following 
propositions. 


PROPOSITION  III. 


Two  Ellipses  having  the  same,  or  equal  Transverse  ax- 
es, are  to  each  other  as  their  Conjugate  axes. 

That  is  Ellip.  Aa  Bh  :   Ellip.  Aa  Bb'y.ab  I  a'h\ 

For  any  ordinate        DE  :  De  :  :CA  :  Ca,  P|;  Vm. 

Therefore  bv  indivis.      CAa  t  CAa':  :Ca  :  Ca'  ;  ^'^'  ^^• 

Or  Ellipse  AaBh  :  Aa'Bb' :  :ab  X  'ah'. 

Q.  E.  IX 

CoR.  1. — This  reasoning  is  equally  applicable  to  two 
Ellipses,  having  the  same  or  equal  conjugate  axes.  They 
are  to  each  other  as  their  transverse  axes. 

That  is,   Ellip.  A'aB'b  :  AaB^>:  :A'B'  :  AB. 

Cor.  ^. — Any  Ellipses  have  to  each  other  the  ratio 
compounded  of  the  ratios  of  their  two  axes. 

For  Ellipse     Aa'Bb'  :  Ellip.  AaBb:  \a'b'  :  ab, 
and  Ellip.       AaBb  :  Ellip.  A'aB'b'.'.A'B'  :  AB, 
.'.    Ellip.       Aa'Bb'  :  Ellip.   A'aB'b   in   a  ratio    com- 
pounded of  the  ratios  of  a'b'  to  ab,  and  of  A'B'  to  AB.  5.  C* 

CoR.  3. — Similar  Ellipses  have  to  each  other  the  du- 
plicate ratio  which  either  of  their  axes  have. 

That  is  Ellipses  Aa'Bb'  :  A'aB'br.AB^  I  A'B'^  :  '.a'b'^:ab\ 

For(22)AB  :  a'b'WA'B'  :  ab, 

and      AB2  :  a'b'  .  AB:  iA'B'^  :  ab  .  A'B', 
.:.  AB^  :  A'B'= : :  AB  .  a'b'  :  AB'  .a6:  :EHip.  Aa'B6'Ellip. 

A'aB'6. 

*-  Simpson's  Euclid.— See  also  Book  VI,  Prop.  XXIIT. 
16 


118  APPENDIX  TO  CONIC  SECTIONS. 


PROPOSITION  IV. 

Two  Hyperbolas,  having  the  same  or  equal  Transvene 
axes  and  equal  abscisses,  are  to  each  other  as  their  Conju- 
gate axes. 

r>g.  15.      That  is,     ADE  :  ADE':  :Ca  :  Ca'. 

This  is  evident  from  the  method  of  indivisibles,  (See 
Hyp.  II,  Cor.  3.) 


PROPOSITION  V. 

Similar  segments  of  similar  Conic  Sections,  are  to  each 
other  in  the  ratio  compounded  of  the  ratios  of  their  ab- 
PI.  VII.  scisses  and  ordinates. 

Fig:.  1.  ADG  :  adg:  :AD  ,  DG  :  ad  .  dg. 

Fig.  2:         ADE  :  A'Je: :  AD  .  DE  :  Arf  .  de, 
Fig.  3.  ADG  :  adg:  \ AD  .  DG  :  a^/  .  dg. 

For  their  abscisses,  being  divided  into  anequal  number  of 
parts,  these  parts  will  be  to  each  otherasthe  whole  abscisses, 
also  the  corresponding  trapeziums  in  the  inscribed  polygons, 
will  be  to  each  other  in  the  ratio  compounded  of  their  bases 
and  altitudes ;  and  the  whole  polygons,  and  consequently 
the  Conic  Sections  which  are  their  limits,  will  be  to  each 
other  as  the  sums  of  the  bases  and  altitudes,  or  as  the  ab- 
scisses and  ordinates. 

CoR.  1. — Similar  segments  are  to  ^ach  other  as  the 
rectangles  of  the  Transverse  and  Conjugate  axes  of  their 
sections. 

CoR.  ^.-Similar  segments,  are  to  each  other,  in  the  du- 
plicate ratio  of  either  of  their  axes. 

ScHOL. — This  proposition  and  corollaries,  follow  from 
Prop.  X^  Cor.  3,  Append.  I. 


APPENDIX  TO  CONIC  SECTIONS.  H9 

PROPOSITION  VI. 

Any  diameter  of  any  Conic  Section,  bisects  any  seg- 
ment included  between  a  double  ordinate  or  the  conju- 
jrate  diameter,  and  the  curve. 


1.  In  the  Parabola  HA  and  HV,  bisect  the  segments  ^'s-  ^2. 
AlandKVI.  '""^ '^' 

Or     KAH=HAI,andKVH=HVI.  Fig.  3. 


2.  In  the  Ellipses  AC,  bisects  aAb  and  EAG, 

Or     ACa=AC6,  and  ADE=ADG. 

also    CG^  =  CE^=CEe  =  CGe. 

The  same  reasoning  evidently  applies  to  the  circle. 


Fig.  13. 


In  the  Hyperbola  AD  bisects  the  segment  EAG,   or  ^»g-  ^' 
ADE-=ADG. 

Also  in  a  triangle,  whether  isosceles  or  scalene,  a  line 
which  bisects  the  base,  bisects  the  triangle. 

For  in  each  case  the  diameter  bisects  the  double 
ordinate  which  limits  the  segment,  and  all  double  ordin- 
ates  within  the  segment,  therefore  by  the  method  of  in- 
divisibles, it  bisects  the  space  in  which  they  are  drawn. 

Cor.  1. — The  spaces  included  between  the  curve  and 
two  tangents  drawn  from  the  extremities  of  any  double  or- 
dinate, are  bisected  by  the  diameter  of  that  double  ordin- 
ate. 

That  is  CAT=TAD,  TNC=TNL  and  TAE=TAH.  Fiff.  i^ 

Also  the    space  between  the  curve,   the  tangent  and  tP.     7* 

double  ordinates  produced  are  equal.  ^  of  EU 

That  is  COP  =  LiR,  FEI  =  GHM,  also  AKE=AK/i.  &  Hyp.' 

CoR.  2.  — The  spaces  between  the  curve  of  an  Ellipse, 
and  a  circumscribing  parallelogram,  whose  sides  are  par- 
allel to  two  coujugate  axes,  are  all  equal. 

That  is,  EKG=s^SG=EQe=Gfe.  ^'^S-  13. 

This  is  equally  applicable  to  the  circle. 


120  APPENDIX  TO  CONIC  SECTIONS. 

Cor.  3. —  The  spaces  between  the  curve  of  a  Hyper- 
bola and  its  asymptotes,  and  bounded  by  a  double  or- 
dinate, are  bisected  by  the  diameter  of  that  double  or- 
dinate. 

That  is  CAEH=CAE^. 


PROPOSITION  VIII. 

Similar  Conoids,  are  to  each  other,  in  the  ratio  com- 
pounded  of  the  ratios  of  their  bases  and  altitudes,  or,  of  the 
squares  of  the  diameters  of  their  bases  and  altitudes. 

Similar  Segments,  being  supposed  to  revolve  on  their 
axes,  generate  similar  Conoids,  and  if  the  inscribed  poly- 
gons, of  which  they  are  limits,  be  supposed  to  revolve  with 
them,  it  will  generate  a  solid,  of  which  the  Conoid  is  the 
limit.  Hence,  by  Indivisibles  or  Exhaustions,  the  proposi- 
tion may  be  deduced. 

Cor.  1. — Similar  Conoids  are  to  each  other,  in  the  ra- 
tio compounded  of  the  ratios  of  the  squares  of  the  diame- 
ters of  their  bases  and  their  altitudes,  or  of  the  squares 
of  their  revolving  axes  or  ordinates,  and  their  fixed  axes 
or  abscisses. 

CoR.  2. — Similar  Conoids  are  to  each  other,  in  the 
triplicate  ratio  of  their  altitudes,  or  the  diameters  of  their 
bases. 


PROPOSITION  IX. 

1.  An  OblateSpheroid,  or  one  described  about  the  Con- 
jugate axis,  is  to  a  Prolate  Spheroid,  or  one  about  the 
Transverse  of  the  same  Ellipse,  as  the  Transverse  axis 
is  to  the  Conjugate  axis. 

2.  The  circumscribing  sphere,  is  to  the  oblate  El- 
lipsoid, as  the  Transverse  axis  is  to  the  Conjugate. 

3.  The  Prolate  Ellipsoid  is  to  the  inscribed  sphere, 
as  the  Transverse  axis,  is  to  the  Conjugate. 


APPENDIX  TO  CONIC  SECTIONS.  121 

4.  The  circumscribing  sphere,  is  to  the  oblate  Ellipsoid, 
as  the  prolate  Ellipsoid  is  to  the  inscribed  sphere.  That 
is,  the  four  solids  are  Geometrical  Proportionals. 

5.  They  are  continued  proportionals,  each  having  to  the 
following,  the  ratio  of  the  Transverse  axis  to  the  Conju- 
gate 5  therefore  the  circumscribing  sphere  has  to  the  in- 
scribed sphere,  the  triplicate  ratio  of  the  Transverse  axis 
to  the  Conjugate,  as  is  evident  also  from  Euclid,  Sup. 
Ill,  n-21. 


PROPOSITION  X. 

Every  Paraboloid,  is  equal  to  one  half  its  circumscri- 
bing cylinder. 

This  is  proved,  in  a  manner  very  similar  to  that  in 
which  the  corresponding  proposition  concerning  the  sphere 
is  demonstrated.     (Plate  XVI.  9) 


PROPOSITION  XL 

Every  Ellipsoid,  Prolate  or  Oblate,  is  f  of  its  cirt:um- 
scribing  cylinder. 

The  Ellipsoid  and  the  sphere,  have  evidently  the  same  pj  yjjj 
relations  to  the  circumscribing  cylinders.  Fig.  12! 


SPHERICAL  GEOMETRY. 


DEFINITIONS  AND    SECTIONS    OF    THE 
SPHERE. 


Def.  1. — A  Sphere  is  a  solid,  generated  by  the  revolu-  g^"*^^ 
tion  of  a  semi-circle  about  its  diameter,  which  remains  ^^{[^^ 
fixed. 


2. — The  Axis  of  the  sphere,  is  the  fixed  diameter  about  ^ 
which  the  semi-circle  is  supposed  to  revolve. 


Def.  9. 


3. — The  Centre  of  a  sphere  is  the  same  as  that  of  the 
semi-circle,  by  which  it  is  described. 

4. — The  Radius  of  the  sphere,  is  any  straight  line, 
drawn  from  the  centre  to  the  circumference  of  the 
sphere. 

Cor.  1. — The  Radii  of  the  sphere  are  all  equal;  each 
being  equal  to  the  radius  of  the  generating  semi-circle. 
In  other  words,  every  point  on  the  surface  of  the  sphere  is 
equally  distant  from  its  centre. 

CoR.  2. — Hence  if  two  portions  of  the  surface  of  the 
same  sphere  or  of  equal  spheres,  be  placed  upon  each 
other  they  will  coincide  in  one  superfices. 

Def.  5.— The  Z)iawc/er  of  the  sphere,  is  any  straight 
line  which  passes  through  the  centre,  and  is  terminated  g^     .^ 
both  ways  by  the  surface  of  the  sphere.  Def/**ia 


124  SPHERICAL  GEOMETRY. 

Cor.— The  diameters  of  a  sphere  are  all  equal ;  being 
each  equal  to  twice  the  radius,  or  to  the  diameter  of  the 
generating  semi-circle. 

Any  diameter,  therefore,  may  be  assumed,  as  the  axis  of 
the  sphere. 

6.— *The  Poles  of  the  sphere,  are  the  extremities  of  iti 
axis. 


Cor. — The  extremities  of  any  diameter  may  be  assum- 
ed as  the  poles  of  the  sphere.     (Def.  5,  Cor.) 

7. — A  plane  touches  the  sphere  when  it  meets  the  sur^ 
face  of  the  sphere,  but  doei  not  cut  it. 

8. — A  straight  line  touches  the  sphere  when  it  meets 
the  surface,  but  does  not  cut  it. 


PROPOSITION  L 

If  a  plane  cut  the  sphere  passing  through  its  centre, 
the  mutual  intersection  of  the  sphere  and  plane,  will  be 
a  circle,  which  is  equal  to  that  by  which  the  sphere  was 
described. 

This  is  evident  because  every  point  of  the  surface  of  the 
•phere,  (Def.  4,  Cor.  1.)  is  equally  distant  from  the  cen- 
tre ;  and  this  distance  is  equal  to  the  radius  of  the  genera- 
1  Def.i2.ting  semi-circle. 

CoR.  1. — All  circles  on  the  sphere  whose  planes  pass 
through  the  centre,  are  equal  to  each  other. 

Scholium. — Circles  on  the  sphere  whose  planes  pass 
through  the  centre,  are  called  Great  Circles. 

Such  upon  the  Armillary  sphere,  are  the  Equator,  the 
Horizon,  the  Ecliptic  and  the  Meridians. 

Cor.  2. — A  great  circle  may  pass  through  any  two  points 
in  the  surface  of  the  sphere. 


SPHERICAL  GEOMETRY.  125 

For  a  plane  may  be  supposed  to  pass  through  arty  three 
^ii-en  points.  As  the  plane  of  a  great  circle  passes  through 
the  centre  of  the  sphere,  it  may  also  pass  through  any 
two  points  in  its  surface. 

Cor.  3. — The  straight  line  which  connects  any  two 
points,  in  the  surface  of  the  sphere,  falls  wholly  within  the 
sphere. 

For  it  is  a  chord  of  the  great  circle  which  passes  through 
these  two  points. 

Cor.  4.  — Any  straight  line  which  touches  the  sphere, 
touches  the  great  circle  whose  plane  coincides  with  the 
line.     (Dcf.  8.)  3  Dei.  h 

Cor.  5. — The  straight  line  which  passes  through  the 
point  of  contact  and  the  centre  of  the  sphere,  is  perpen- 
dicular to  a  line  touching  it. 

In  other  words,  any  diameter  of  the  sphere,  is  perpen- 
dicular to  all  lines,  which  meet  its  extremity  and  touch  the 
sphere.  3,  ig, 

CoR.  6.  — An  indefinite  number  of  great  circles  may 
pass  through  any  given  point  in  the  surface  of  the  sphere. 

For  planes  may  pass  through  the  centre,  through  a  giv- 
en point  in  the  surface,  and  through  any  other  point  in  the 
surface.  And  the  intersection  of  each  of  them  with  the 
sphere  will  be  a  great  circle.     (Cor.  2.) 

ScHOL.  -  Thus  there  are  an  indefinite  number  of  Me- 
ridians, all  passing  through  the  poles  of  the  terrestial 
sphere. 

Cor.  7.— The  mutual  intersections  of  a  plane  which 
touches  a  sphere,  and  the  planes  of  any  number  of  great 
circles  passing  through  the  point  of  contact,  will  be 
straight  lines  touching  the  sphere,  (Def.  7.)  and  touching 
those  great  circles  respectively.  (Cor.  4.) 

17 


Sup.  2. 
Dei;    I. 


126  SPHERICAL  GEOxMETRY. 

ScHOL.  —  Thus  if  a  plane  touch  the  globe  at  one  of  its 
poles,  the  mutual  intersections  of  this  plane  with  the 
planes  of  the  Meridians,  will  all  be  lines  touching  those 
Meridians,  and  also  touching  the  sphere. 

Cor.  8.  — The  radius  or  diameter  of  the  sphere,  which 
passes  through  the  point  of  contact,  is  perpendicular  to 
\\\ii  plane  touching  the  sphere. 

For  it  is  perpendicular  to  every  straight  line  which  it 
meets  in  that  plane.  (Cor.  5,  and  7.) 

ScHOL.  — Thus  the  axis  of  the  Earth,  is  perpendicular 
to  planes  touching  it,  at  its  poles. 

CoR.  9.  —  The  mutual  intersection  of  two  great  circles, 
is  a  diameter  of  the  sphere. 

For  as  the  plane  of  each  great  circle  passes  through  the 
centre  of  the  sphere,  the  line  of  their  mutual  intersection 
ihcrtfore  passes  through  the  centre  of  the  sphere.  It  is 
therefore  a  diameter.     (Def.  5.) 

CoR.  1 0.  —  Hence  all  great  circles  mutually  bisect  each 
other  :  that  is,  they  divide  each  other  into  two  equal  parts, 
or  semi-circles. 

ScHOL.— Thus  the  Ecliptic  and  the  Equator,  bisect 
each  other.     All  the  Meridians  bisect  each  other,  &c. 


PROPOSITION  II. 


If  any  plane  cut  a  sphere,  but  do  not  pass  through  the 
centre,  the  section  will  be  a  circle,  whose  radius  is  equal 
to  that  ordinate  in  the  generating  semi-circle  which  coin- 
cides with  the  plane. 

For  in  describing  the  sphere,  the  revolutions  of  that 
ordinate  will  describe  the  circle. 

CoR.  1. — The  circles  formed  by  the  section  of  planes 
which  do  not  pass  through  (he  centre  of  the  sphere,  since 
they  are  to  each  other  as  the  squares  of  their  diameters  or 
radii, — that  is,  the  squares  of  the  ordinates  in  the  semi- 


SPHERICAL  GEOMETRY.  127 

circle,  are  as  the  rectangles  of  the  segments  into  which 
their  planes  divide  the  axis. 

That  is,  the  cir.  EcQ  :  HDI:  :PC  .  C/>  :  PD  .  Dp.        Fig;,  i- 

ScHOL. — As  these  circles  are  smaller  than  the  genera- 
ting circle  of  the  sphere,  and  therefore  smaller  than  the 
circular  sections  which  pass  through  the  centre,  they  are 
called  Small  Circles  of  the  sphere.  Such  are  all  the 
allels  of  latitude  on  the  terreatial  sphere. 


uli- 


CoR.  2. — Small  circles  whose  planes  are  equally  distant 
from  the  centre,  are  equal.  •' 

And  conversely. — If  they  are  equal,  they  are  equally 
distant  from  the  centre. 

For  then,  the  corresponding  rectangles,  of  the  axis,  will 
be  equal.     (Cor.  1.) 

Cor.  3. — If  two  small  circles  are  unequal,  the  less  is 
more  distant  from  the  centre,  and  conversely. 

ScHOL. — Thus  the  two  Tropics,  or  the  Polar  Circles, 
are  respectively  equal,  and  equally  distant  from  the  cen- 
tre ;  but  the  Polar  Circles  are  more  distant  from  the 
centre  than  the  Tropics,  and  smaller. 

Cor.  4. — A  small  circle  may  pass  through  any  three 
points,  in  the  surface  of  the  sphere,  if  those  points  are  not 
all  in  the  same  great  circle. 

For  a  plane  may  pass  through  ani/  three  points,  and  if  it 
do  not  pass  also  through  the  centre  of  the  sphere,  the  sec- 
tion will  be  a  small  circle. 

Cor.  5. — If  any  two  circles  meet,  but  do  not  cut  each 
other,  on  the  surface  of  the  sphere,  the  straight  line  which 
touches  one  of  the  circles,  touches  the  other  also. 

For  it  meets  the  other  circle,  but  does  not  cut  it. 
If  the  line  did  cut  the  second  circle,  it  would  fall 
wholly  within  the  sphere  (I,  Cor.  2.)  and  therefore 
would  not  touch  the  first  circle,  contrary  to  the  suppo- 
sition. 


128  SPHERICAL  GEOMETRY. 


1 


And  conversely,  Circles  which  touch  the  same  straight 
line  touch  each  other. 

Cor.  6. — Hence  the  mutual  intersection  of  the  planes  of 
two  circles  which  touch  each  other  on  the  sphere,  is  a 
straight  line  which  touches  both  of  them. 

For  the  line  is  in  the  planes  of  both  the  circles,  and  it 
touches  each  of  them. 


ScHOL. — As  illustrations  :  The  Tropics  touch  the 
Ecliptic.  Each  of  them  therefore,  and  the  Ecliptic  touch 
the  same  straight  line,  at  the  point  of  contact,  and  that 
line  is  the  common  intersection  of  their  planes. 

Def.  9. — The  extremities  of  the  axis,  are  the  poles  of 
all  circles  which  are  perpendicular  to  the  axis. 

Cor.  1. — Hence  two  great  circles  cannot  have  the  same 
poles,  not  having  a  common  axis. 

Cor,  2.  — There  may  be  an  indefinite  number  of  small 
circles,  all  parallel  to  a  great  circle,  and  having  the  same 
poles  with  it,  being  all  perpendicular  to  the  same  axis. 

ScHOL. — Thus  all  the  parallels  of  latitude  have  the 
same  poles  as  the  Equator. 

CoR.  3. —  Circles  which  have  the  same  poles  are  par- 
allel to  each  other,  and  the  axis  passes  through  the  cen- 
tre of  each  of  them. 

The  axis  therefore  passes  through  the  poles  of  any  cir- 
cle, through  its  centre,  and  through  the  centre  of  the 
sphere. 

CoR.  4. — The  plane  of  any  great  circle,  is  at  right  an- 
"^'  '     gles  to  all  the  circles  through  whose  poles  it  passes.   For  it 
coincides  with  the  axis,  which  (Def.  9.)  is  perpendicular  to 
all  their  planes. 

ScHOL. — Thus  the  Meridians  are  at  right  angles  to  the 
Equator  and  to  all  its  parallels. 


SPHERICAL  GEOMETRY.  129 

Cor.  5. — And  conversely,  All  the  great  circles,  which 
are  at  right  angles  to  a  given  great  circle,  intersect  each 
other  in  its  poles. 

For  they  all  pass  through  its  poles,  and  must  therefore 
intersect  each  other  there. 

ScHOL.  — For  example,  all  the  Meridians  are  at  right 
angles  to  the  Equator,  and  they  all  intersect  each  other  in 
its  poles. 

CoR.  6.  —  So  also,  if  two  great  circles  are  at  right  angles 
to  each  other,  the  poles  of  each  are  in  the  circumference 
of  the  other :  and  if  any  number  of  great  circles  pass 
through  the  poles  of  another  great  circle,  the  poles  of  all 
the  former,  are  in  the  circumference  of  the  latter. 

ScHOL. — Thus  the  poles  of  all  Meridians,  are  in  the  E- 
quator. 

Cor.  7. — If  three  great  circles  are  at  right  angles  to 
each  other,  the  pole  of  each  is  in  the  circumference  of 
both  the  others,  and  therefore  in  the  point  of  their  inter- 
section. 

CoR.  8.  — A  circle  whose  plane  coincides  with  the  axis, 
and  therefore  passes  through  the  poles,  bisects  all  the  cir- 
cles which  are  perpendicular  to  the  axis. 

For  it  passes  through  their  centres.     (Cor.  3.) 

CoR.  9. — The  angle  made  at  the  centre  of  the  sphere, 
by  the  axes  of  two  great  circles,  is  equal  to  the  inclinatioi 
of  their  planes. 

That  is  PCP=QCL, 

For  PCQ=P'CL=  a  right  angle,  therefore  taking  away  ^»S-  2 
the  common  part  KCQ,  and  PCP'=QCL. 

CoR.  10. — If  two  circles  touch  each  other,  on  the  sur- 
face of  the  sphere,  the  piano  which  passes  through  the 
point  of  contact,  and  is  perpendicular  to  the  line,  touching 
both  of  them.  (11  Cor.  5.)  that  is,  to  the  line  of  their  mu- 


]3()  SPHERICAL  GEOMETRY. 

Sup.2.i7.tual  intersection,  (II  Cor.  6.)  is  perpendicular  also  to 
their  planes.  It  therefore  passes  through  their  poles, 
(Cor.  5.)  and  through  the  centre  of  the  sphere, 
(Cor.  3.) 

Conversely,  The  plane  of  a  great  circle  which  passes 
Sup.2.l8.through  the  poles  of  two  other  circles  which  touch  each 
other,  passes  also  through  the  point  of  contact  and  is  per- 
pendicular to  the  line  of  their  mutual  intersection. 

ScHOL. — For  example.  The  solstitial  Colure.  passes 
through  the  points  of  contact,  of  the  Ecliptic  and  Tropics, 
and  through  their  poles.  It  is  perpendicular  also  to  their 
planes,  and  to  the  lines  of  their  muttil  intersection,  each 
of  which  lines  touches  both  the  Ecliptic  and  Tropic  in  the 
points  of  their  contact. 

CoR.   11. — Any    circle  on    the   sphere  cuts  off   equal 
arcs   from    all  the  great  circles   which  pass   through  its 
Fig.  2.  poles. 

That  is,  PH=PF  and  PE=PG. 

For  in  the  description  of  the  sphere,  the  arc  of  the  gen- 
erating semi-circle  between  the  poles,  and  any  plane  per- 
pendicular to  the  axis,  will  coincide  successively  with  the 
arcs  of  all  great  circles  passing  through  the  pole. 

ScHOL.— Thus  the  Equator,  or  any  of  its  parallels,  cuts 
off  equal  arcs  from  all  the  Meridians. 


Cor.  12. — Hence  any  two  parallel  circles,  intercept 
equal  arcs  of  all  great  circles  passing  through  their 
poles. 

That  is     EH=GF. 

This  is  inferred  from  the  last,  by  taking  equal  arcs  from 
those  which  are  equal. 


SPHERICAL  GEOMETRY.  13^ 

ScHOL. — Thus  arcs  of  Meridians,  intercepted  between 
two  parallels  of  Latitude,  or  between  any  parallel  and  the 
Equator,  are  equal. 


PROPOSITION  III. 

If  two  circles  on  the  sphere,  pass  through  the  remotest 
poles  of  two  great  circles,  they  will  cut  oif  equal  arcs  from 
those  circles. 

That  is  Ke=Gg,  Fig.  3. 

Forthe  arcPE=;}G;  .'.  PG=;oE, 

Therefore  the  chord  PG  =pE,  and  PD=;?D.  3. 3C. 

A^ain  the  chord  Fe=pg^  .',  arc  Fe=^pg. 
Therefore  the  arc  Pg=pe  and  chord,  Fg=pe,  and  Vd^=pd^ 
Therefore,  the  triangle  DPc/=D/)c/  in  all  respects;  i-  8- 

.'.  also  GPg  =  E/?e,  and  chord  gG=Ee;  and  arc  Gg^=Ee. 

Q.  E.  D. 

PROPOSITION  IV. 

If  from  any  point  which  is  not  the  pole  of  a  great  cir- 
cle, there  be  described  arcs  of  great  circles  to  that  circle, 
the  greatest  is  that  which  passes  through  its  pole.  And 
the  other  arc  of  the  same  circle  is  the  least;  and  of  the 
others,  that  which  is  nearer  to  the  pole,  is  greater  than 
any  other  which  is  more  remote. 

Let  the  common  section  of  the  planes  of  the  great  cir-  Fij.  7. 
cles  ACB,   ad  B  at  right  angles  to  each  other,   be  AB; 
and  from  C,  draw  CG  perpendicular  to  AB,  which  will  ^jj;  ^ 
also  be  perpendicular  to  the  plane  ADB, 

Jom  GD,  GE,    GF,  CD,  CE,  CF,  CA,  CB. 


Of  all  the  straight  lines  drawn  from  G,  to  the  cir- 
cumference ADB,  GA  is  the  greatest  and  GB  the  least ; 
and  GD  which  is  nearer  to  GA  is  greater  than  GE, 
which  is  more  remote.  The  triangles  CGA,  CGD  are 
right  angled  at  G,  and  they  have  the  common  side  CG  ; 
therefore  the  squares  of  CG,  GA  together,  that  is  the 
square  of  CA,  is  greater  than  the  squares  of  CG,  CD  to- 
gether, that  is.  the  square  of  CD  :  and  CA  is  greater  than 


3.  7 


^32  SPHERICAL  GEOMETRY. 

a  28.  CD,  and  therefore  the  arc  CA  is  greater  than  CD.  In 
the  same  manner,  since  GD  is  greater  than  GE,  and  GE 
than  GF,  &;c.  it  is  shown  that  CD  is  greater  than  CE,  and 
CE  than  CF  he,  and  consequently,  the  arc  CD,  greater 
than  the  arc  CE,  and  the  arc  CE  greater  than  the  arc  CF, 
&c.  And  since  GA  is  the  greatest,  and  GB  the  least  of  all 
the  straight  lines  drawn  from  G  to  the  circumference  ADB, 
it  is  manifest  that  CA  is  the  greatest,  and  CB  the  least  of 
all  the  straight  lines  drawn  from  C  to  the  circumference  : 
and  therefore  the  arc  CA  is  the  greatest,  and  CB  the 
least  of  all  the  circles  drawn  through  C,  meeting  ADB. 

Q.  E.  D, 


SPHERICAL  GEOMETRY. 


II.  OF  SPHERICAL,  TRMJVGLES. 


Def.  i— a  spherical  angle  is  the  angle  made  by  two 
circles  which  intersect  each  other,  on  the  surface  of  the 
sphere.  Or  it  is  the  angle  made  by  the  straight  lines 
7vhich  touch  those  circles  at  the  point  of  their  intersection* 

ScHOL. — The  first  of  the  preceding  definitions,  is  one 
frequently  given  by  writers  on  spherics.  '*  The  angle 
made  by  tiuo  circles^^^  however  is  two  indefinite  an  expres- 
sion. CagnoH  endeavors  to  render  it  more  precise,  by 
adding  that  ?7  is  the  angle  made  by  the  arcs,  "  considered  in 
the  points  immediately  contiguous  to  that  in  which  they  meet 
each  other**''  (consideree  dans  les  points  immediatement 
contigiis  a  celui  dans  lequel  Us  se  rencontrent^  (279.)  and 
this  angle  he  infers  is  the  same  as  that  made  by  the  lines 
touching  them  in  that  point.  But  not  to  mention  the  want 
of  mathematical  precision  in  the  phrase  "points  imme- 
diately contiguous  to  that  <Sz:c,"  the  equality  of  this  angle 
and  that  of  the  touching  lines,  cannot  be  rigorously  de- 
monstrated except  by  the  method  of  ultimate  or  limiting 
ratios,  a  method  far  too  refined  and  too  far  removed  from 
the  principles  of  Elementary  Geometry  to  be  involvcjd  in 
a  simple  definition.  To  avoid  this  difficulty  the  second 
of  the  preceding  definitions  is  given,  which  will  be  refer- 
red to  in  the  following  demonstrations. 

CoR.  1. — Hence  if  two  circles  cut  each  other  on  the 
surface  of  the  sphere,  the  adjacent  angles  are  either  two 
right  angles,  or  are  together  equal  to  two  right  angles,  i.  is. 

That  is  BCA-f-BCD  =  ECA+ECD=2  right  angles,     fig.  lo. 

1& 


134  SPHERICAL  GEOMETRY. 

For  this  is  true  of  the  angles  made  by  the  straight  lines 
touching  them  at  the  point  of  intersection. 

Cor.  2. — For  the  same  reason,  the  angles  made  by  any 
two  circles,  or  any  number  of  circles  intersecting  each 
other  in  one  point,  on  the  surface  of  the  sphere,  are  to- 
gether equal  to  four  right  angles. 

Cor.  3. —  A  spherical  angle  is  always  less  than  two 
right  angles. 

J   j5       Cor.  4. — If  two  circles  cut  each  other  on  the  sphere, 
the  vertical  or  opposite  angles  are  equal. 

Fig.  11.  That  is  BCA  =  ECF. 

Cor.  5.  — The  angle  which  two  great  circles  make  up* 
on  the  surface  of  the  sphere,  is  the  same  as  the  inclina- 
tion of  the  planes  of  these  circles. 

Fig.  2.  That  is  GPE=GCE. 


For  the  lines  touching  the  circles  in  the  point  of  inter- 
section, are  perpendicular  to  that  diameter  of  the  sphere, 
which  is  the  line  of  the  mutual  intersection  of  their  planes  ; 
(I    Cor.   5.)    the  angle    formed  by    the   touching    lines 

Sap.  2.  therefore,  is  the  inclination  of  their  planes.     It  is  also  the 

Def.  4.  angle  made  by  the  circles.     (Def.  1.) 

Cor.  6, — Hence  the  angles  made  by  any  two  great 
circles,  on  the  opposite  side  of  the  sphere  are  equal. 
That  is  GPE  =  G/?E. 

Cor.  7. — The  angles  made  by  two  great  circles  are 
equal  to  that  made  by  two  of  their  radii,  which  are  in  the 
plane  of  another  great  circle  whose  pole  is  the  angular 
point. 

For  these  two  radii  are  parallel  to  the  two  lines  touch- 
ing the  circles  in  the  angular  point.     (See  Cor.  5.) 


SPHERICAL  GEOMETRY.  13*, 

CoK.  8. —  Hence  the  angles  which  great  circles  make 
with  each  other  on  the  surface  of  the  sphere,  are  propor- 
tional to  the  arcs,  intercepted  by  them,  of  a  great  circle 
whose  pole  is  the  angular  point. 

For  EG  is  the  measure  of  GCE,  therefore,  Fig;  2. 

it  is  also  the   measure   ofGPE.     (Cor.  7.) 

ScHOL. — Thus  any  arc  of  the  Equator  intercepted  be- 
tween two  Meridians,  is  the  measure  o(  the  an^le  which 
those  meridians  make  at  its  poles.  Whenever  spherical 
angles  are  compared  with  arcs  of  great  circles,  the 
measures  of  these  angles,  are  really  compared  with  those 
arcs. 

Cor.  9. — The  arcs  of  parallel  circles,  intercepted  by 
great  circles  which  pass  through  their  common  poles,  (9 
Cor.  2.)  arc  to  each  other  as  their  circumferences,  (since 
they  subtend  equal  angles  at  their  centres,)  that  is  as  their 
diameters  or  radii.  Therefore  the  arcs  oi  any  circle  may 
be  made  the  measures  of  the  angles  formed  by  great  cir- 
cles at  its  poles,  since  they  are  always  proportional  to 
those  angles. 

That  is  FH  is  the  measure  of  FDH=GCE  =  GPE. 

Def.  2.  — a  Spherical  Triangle,  is  a  figure  formed  on 
the  surface  of  the  sphere,  by  the  arcs  of  three  great  cir- 
cles, which  intersect  each  other. 

Cor.  1.  — The  side  of  a  spherical  triangle  is  always 
less  than  a  semi-circle. 

For  any  two  great  circles  intersect  each  other  in  points 
diametrically  opposite,  dividing  each  other  into  semi-cir- 
cles. (1,  Cor.  10.)  Therefore  if  a  third  circle  intersect 
each  of  these,  it  must  intersect  the  semi-circles  into  which 
they  divide  each  other,  and  consequently  each  intercept- 
ed arc  will  be  less  than  a  semi-circle. 

That  is,  PH^  and  PEp  are  semi-circles;  therefore  iff'ig-^. 
any  arc,  as  HF  intersect  them,  forming  the  triangle  ^HF, 
/)H,  or  pF,  are  each  less  than  ^HP,  or  ^FP,   and  there- 
fore less  than  a  semi-circle. 


136  SPHERICAL  GEOMETRY. 

Def.  3. — The  different  kinds  of  Spherical  Triangles.  — 
right-angled^  acute-angled^  obtuse-angled^  and  equi-angu- 
lar  ;  also  equi-lateral,  isoceles,  and  scalene,  are  distinguish- 
ed and  defined  in  the  same  manner  as  in  plane  triangles. 
(See  Euclid,  Book  I,  Def.  20  to  25.) 

Def.  4. —A  rectilateral,  or  quadrantal,  spherical  tri- 
angle, has  one  side  or  quadrant. 

Def.  5. — An  oblique  angled  triangle^  is  a  general 
term  applied  to  all  triangles  in  which  there  is  not  a  right- 
angle,  nor  a  quadrantal  side. 


PROPOSITION  I.   PROB. 

On  a  given  arc,  to  describe  (on  the  surface  of  the 
sphere)  an  Equilateral  Spherical  Triangle. 

Let  AB  be  the  given   arc,  on  which  it  is    required    to 
Eig.  5.   describe  an  Equilateral  Spherical  Triangle. 

Let  one  extremity  of  the  arc  as  A,  be  taken  as  a  pole, 
and  through  the  other  extremity,  as  B,  let  a  small  circle 
be  described  whose  pole  is  A,  (Def.  9,  Cor.  2,  also  II,  2,) 
then  let  B  be  taken  as  a  pole,  and  a  small  circle  described 
about  it  passing  through  A,  they  will  intersect  each  other 
in  C.  Then  let  two  great  circles  pass  through  A  and  C, 
and  through  B  and  C  (I,  Cor.  2,)  ABC  is  the  Triangle  re- 
quired. 

For  AC=AB  (II.  Cor.  11,)  and  BC=BA.-.AC=BC-= 
AB. 

CoR.  —Therefore,  from  any  given  point  on  the  surface 
of  the  sphere,  an  arc  of  a  great  circle  may  be  described 
equal  to  a  given  arc,  in  a  manner  analogous  to  that  follow- 
ed in  the  second  proposition  of  the  first  book  of  Euclid  ; 
for  a  small  circle  cuts  off  equal  arcs  of  great  circles  pass- 
ing through  its  poles  ;  as  in  a  plane,  a  circle  cuts  off  equal 
segments  from  lines  passing  through  its  centre. 


SPHERICAL  GEOMETRY.  J37 


PROPOSITION  II. 


If  two  gpherical  triangles  have  two  sides  of  the  one, 
equal  to  two  sides  of  the  other  each  to  each,  and  the  con- 
tained angles  equal,  the  triangles  are  equal  in  all  re- 
spects ;  that  is,  their  areas  are  equal,  the  remaining  sides  in 
each  are  equal,  and  the  angles  opposite  the  equal  sides  are  , 

equal. 

If  the  two  equal  sides  in  one,  correspond  io  those  in  the  ^''o-  ^0 
other  in  position,  as  in  the  triangles  CAE  and  CAB,  their  ^'^^^'^ 
equality  may  be  proved  in  the  method  followed  by  Euclid 
Prop.IV,  by  super-position, and  coincidence  (4. Def.Cor.2.) 

If  they  do  not  correspond  in  position,  as  CGt  and  CGF,  '^'o-U 
supposing  that  CE=CF,  and  GE=GF  it  is  evident  they  can- 
not be  made  to  coincide ;  it  is  then  perhaps  a  sufficient  reason 
for  believing  them  equal,  that  there  is  no  reason  why  they 
should  differ.  Their  equality  however  may  be  rigorous- 
ly demonstrated.  * 


PROPOSITION  III. 

If  iwo  spherical  triangles,  have  one  side  in  each  equal, 
and  the  angles  adjacent  to  the  equal  sides  equal,  each  to 
each,  the  two  triangles  shall  be  equal  in  all  respects.  Fig.  10. 

That  is  if  ACE  and  ACB  have   the   side  AE   in  one,  a^^  12- 
equal  to  AB  in  the  other,  and  CAE,  CEA,  in  one,   equal 
CAB,  CBA,  in  the  other,  then  AC=AC   and   CE=CB 
and  ACE  =  ACB. 

This  proposition  may  be  demonstrated  by  super-posi- 
tion, as  Prop.  II  ;  or  it  may  be  demonstrated  by  a  reductio 
ad  ahsurdum  founded  on  Prop.  II,  as  Euclid  has  done  in 
regard  to  plane  triangles.     (Book  I,  26,  first  part.) 


PROPOSITION  IV. 

The  angles  at  the  base  of  an  isosceles  spherical  triangle 
are  equal. 

That  is,  if  CE=CF,  then  CEF=CFE.  Fig  11. 

^  See  Notes. 


138  SPHERICAL  GEOMETRY. 

This  proposition  may  be  demonstrated  from  Prop.  II,  in 
a  manner  similar  to  that  in  which  the  corresponding  prop- 
osition is  demonstrated  in  plane  triangles.  (See  Euclid 
I  Prop.  5.) 

1.5.  Cor.     Cor.   1. — Hence  an   Equilateral  Spherical  Triangle,  is 
also  Equiangular. 

Cor.  2.  —  And  conversely,  if  the  angles  at  the  base  of  a 
spherical  triangle  are  equal,  the  opposite  sides  are  also 
equal. 

This  is  derived  from  the  Proposition  and  may  be  de- 
monstrated by  reductio  ad  absurdum  in  the  manner  of  Eu- 
clid Prop.  VI. 

Cor.  3. — Hence,  every  Equiangular  Spherical    Trian- 
l.e.Cor.  gle,  is  also  Equilateral. 


PROPOSITION  V. 

If  two  Spherical  Triangles  have  the  three  sides  of  one 
equal  to  the  three  sides  of  the  other,  tbe  two  triangles  will 
be  equal  in  all  respects. 

This  proposition  may  be  demonstrated  by  proving  that 
upon  the  same  base,  and  upon  the  same  side  of  it,  two 
spherical  triangles  cannot  be  described,  which  have  their 
I.  7.     corresponding  sides  equal,  as  is  done  in  plane  triangles. 

ScHOL.  — The  Problems,  (or  bisecting  a  spherical  angle  ^ 
— for  bisecting  a  given  arc  of  a  great  circle,  for  describing 
an  arc  which  shall  be  at  right  angles  to  another  arc,  ei- 
ther from  a  point  within  the  arc,  or  from  a  point  above  it, 
depend,  like  the  similar  Problems  in  plane  triangles,  upon 
describing  an  Equilateral  Triangle,  upon  a  given  base,  and 
drawing  another  arc  through  two  given  points. 

Prop.  I,  describes  an  Equilateral  Spherical  Triangle, 
and  an  arc  of  a  great  circle  may  be  described  through  any 
two  points  on  the  surface  of  a  sphere,  (1,  Cor.  2.)  and  there- 
fore, through  the  vertices  of  two  triangles,  in  the  first  and 
second  cases,  and  through  the  vertex  of  a  triangle,  and  a 
point  in  the  base,  in  the  two  latter  cases. 


SPHERICAL  GEOMETRY.  I39 

Also,  by  the  same  means,  upon  a  given  base,  a  splieric- 
al  triangle  may  be  described,  whose  sides  shall  be  equal  to 
those  of  a  given  spherical  triangle.  And  consequently,  a 
spherical  angle  may  be  described,  equal  to  any  given 
spherical  angle.  (See  Euclid  I,  22,  23.)  Figures  4,  and 
5,  may  be  used  for  the  illustration  and  demonstration  of  all 
the  propositions  referred  to. 


PROPOSITION  VL 

Any  two  sides  of  a  spherical  triangle  are  greater  than 
the  third  side. 

That  is  AB  +  BC>AC,  and  AB-f  AC>BC,  and  BC+AC  Fig.  8. 
>AB. 

For  the  planes  of  the  three  circles  all  pass  through  D,  the 
centre  of  the  sphere,  forming  a  solid  angle  at  D,  any  two 
angles  of  which  are  greater  than  the  third  ;  therefore  the  Sup.  2. 
same  is  true  of  the  arcs  AB,   BC  and  AC  which    have   to     20. 


each  other  the  same  ratios  as  the  angles. 


6.  33. 


Q.  E.  D, 


PROPOSITION  \n. 

The  three  sides  of  a  spherical  triangle,  are  together 
less  than  a  circle. 

For  any  two  great  circles,  as  ACD,  ABD,   bisect   each  Fig. 
other,  therefore  ACD  and  ABC,  are  together  equal  to   a 
circle. 

l^et  the  arc  CB,  intersect  both  these  circles. 

In  the  triangle  CAB  the  two  sides  CA  and  AB,  are  to- 
other greater  than  the  third  side  CB  (VI,)  therefore  CD 
DB  and  BC,  are  together  less  than  CD,  DB,  BA  and  AC, 
that  is  less  than  a  circle. 

Q.  E.  D. 

CoR.  -Each  bf  the  three  sides  of  a  spherical  triangle 
may  be  greater  than  a  quadrant. 

Also  each  of  them,  evidently,  may  be  less  than  a  quad- 
rant. 


10. 


140  SPHERICAL  GEOMETRY. 


PROPOSITION  Vlll. 


In  any  spherical  triangle,  the  greatest  side   is   opposite 
the  greatest  angle  ;  and  conversely. 
Fig.  10.       That  is,  if  the  angle  ACE  is  greater  than  AEC  then  al- 
so AE  is  greater  than  AC. 

For,  let  the  angle  ACB  be  made  equal  to  BAC,  then 
BA=BC  (III)  and  therefore  CB+BE=AE  ;  but  CB-h 
BE>CE,  therefore  AE>CE,  that  is  the  greater  side, 
AE  in  the  triangle  ACE,  is  opposite  to  the  greater  angle  C. 

Q.  E.  D. 


PROPOSITION  IX. 

If  the  three  angles  of  any  spherical  triangle  are  made 
the  poles  of  three  great  circles,  the  intersection  of  those 
circles  will  be  a  triangle,  whose  sides  will  be  supplemen- 
tal to  the  measures  of  the  angles  of  the  given  triangle, 
and  the  measures  of  its  angles  will  be  supplemental  to  the 
sides  of  the  given  triangle. 
Fig  9.  That  is  DF,  FE,  and  ED  are  supplemental  of  the  an- 
gles C,  B  and  A,  and  the  angles  at  E,  D  and  F,  are  sup- 
plemental of  the  arcs  DF,  FE  and  ED. 

Since  A  is  the  pole  of  FE  and  the  circle  AC  passes 
through  A,  EF  will  pass  through  the  pole  of  AC  (9,  Cor. 
6.)  and  since  C,  is  the  pole  of  FD,  FD  also  will  pass  through 
the  pole  of  AC  ;  therefore  the  pole  of  AC  is  in  the  point 
F,  in  which  the  arc  DF,  EF  intersect  each  other.  In 
the  same  manner  D  is  the  pole  of  BC,  and  E  the  pole  of 
AB. 

And  since  F,  E,  are  the  poles  of  AL  and  AM,  FL  and 
EM  are  quadrants,  and  FL,  EM  together,  that  is  FE  and 
ML  together,  are  equal  to  a  semi-circle.  But  since  A  is 
the  pole  of  ML,  ML  is  the  measure  of  the  angle  BAC. (II, 
I  Cor.  8.) 

In  the  same  manner,  ED,  DF  are  the  supplements  of 
the  measures  of  the  angles  ABC,  BCA. 


SPHERICAL  GEOMETRY.  141 

Since  likewise  CN,  BH  are  quadrants,  CN,  BH  togeth- 
er, that  is,  NH,  BC  together  are  equal  to  a  semi-circle  ; 
and  since  D  is  the  pole  of  NH;  NH  is  the  measure  of  the 
angle  FDE  ;  therefore  the  measure  of  the  angle  FDE 
is  the  supplement  of  the  side  BC.  In  the  same  man- 
ner, it  is  shown  that  the  measures  of  the  angles  DEF, 
EFD  are  the  supplements  of  the  sides  AB,  AC,  in  the  tri- 
angle ABC. 

Q.  E.  D. 

ScHOL.  — The  triangle  DEF,  is  called  the  Supplemental 
or  Polar  triangle  of  ACB. 


PROPOSITION  X. 

If  two  triangles  have  the  three  angles  of  one  equal  to 
the  three  angles  of  the  other  each  to  each,  the  two  trian- 
gles shall  be  equal  in  every  respect. 

For  the  sides  of  their  Polar  Triangles  are  then  equal, 
each  to  each  (IX,)  therefore  the  two  Polar  Triangles  shall 
be  equal  to  each  other  in  every  respect,{V.)  Therefore 
the  sides  of  the  give7i  triangles,  which  are  supplemental 
to  the  angles,  respectively,  of  i\\e  polar  triangles,  shall  be 
equal  to  each  other,  each  to  each,  therefore  the  two  given 
triangles  are  equal  in  all  respects.  (V.) 

Q.  E.  D, 


PROPOSITION  XI. 

The  three  angles  of  any  spherical  triangle  are  together 
greater  than  two  right  angles,  and  less  than  six  right  angles. 

That  is  A-j-B-{-C>  2  right  angles  and  <  6  right  angles.  Fig.  9. 

First  they  are  greater  than  two  right  angles,  for  the 
sides  of  its  polar  triangle  DP,  FE  and  ED,  which  are.  sup- 
plemental to  their  angles,  are  together  less  than  a  circle 
(VII)  which  is  the  measure  of  four  right  angles,  therefore 
their  three  supplemental  angles  C,  A,  B  are  greater  than 
two  right  angles. 

19 


^42  SPHERICAL  GEOMETRY. 

Second,  As  any  spherical  angle  (II  Def.  1,  Cor.  3.)  is 
less  than  two  right  angles,  the  three  angles  of  any 
spherical  triangle,  are  together  less  than  six  right  angles. 

Q.  E.  jD. 

ScHOL.  — In  the  property  now  demonstrated,  there  is 
H  striking  difference  between  spherical  and  plane  triangles, 
(see  Euclid  I,  32.)  which  is  the  foundation  of  correspond- 
ing differences  in  all  the  relations  deducible  from  these  dif- 
ferent properties. 

Cor.   1,  — If  a  side  of  a  spherical  triangle  be  produced, 
the  exterior  angle,  is  less  than  the  sum  of  the  two  interior 
and  opposite  angles. 
Fig.  10.  That  is,  CED  is  less  than  ECA  +  EAC. 

For  this  exterior  angle  CED  together  with  its  adjacent 
interior  angle  CEA  is  equal  to  two  right  angles,  but  the 
angle  CEA  together  with  the  two  opposite  interior  angles, 
is  greater  than  two  right  angles. 

Cor.  2. — Also  if  the  sides  of  a  spherical  triangle  be 
produced,  the  exterior  angles  are  together  less  than  four 
right  angles. 

CoR.  3. — The  three  angles  of  a  spherical  triangle,  may 
be  each  greater  than  a  right  angle. 


PROPOSITION  XII. 

In  any  spherical  triangle,  if  the  sum  of  the  tivo  sides,  be 
equal  to  a  semi-circle,  the  external  angle  at  the  base,  will 
be  equal  to  the  interior  and  opposite  angle  ;  and  therefore 
the  sum  of  the  two  angles  at  the  base,  will  be  equal  to  two 

Fi-.  10.  ri?*^^  angles. 

That  is,  if  AC4-CE  =  AD,then  CED=CAE. 
and         CAE-f  CEA=two  right  angles. 

For  ihen    CE=CD.-.(IV)  CED=CDE=CAE. 

And       CAE-f  CEA=CEA4-CED=rtwo  right  angles. 

Q.  E.  D. 


SPHERICAL  GEOMETRY.  j43 

Cor.  1. — Conversely,  if  the  angles  at  the  base,  are  equal 
to  two  right  angles,  the  sides  are  together  equal  to  a  semi- 
circle; and  if  the  angles  are  unequal,  that  opposite  the 
greater  angle,  is  greater  than  a  quadrant,  (Vl[I)and  that  op- 
posite the  less,  is  less  than  the  quadrant :  and  conversely. 
But  if  the  angles  are  equal,  the  angles  at  the  base  are 
right  angles. 
That  is,  if  CAEH-CEA=2  right  angles,  then  AC4-CE  = 

AD. 
And  if    CAE  =  CEA,   then  CE  and  CA,   each   equal  a 
quadrant. 
If        CAE>CEA      "     CE  is  greater,  and  CA  less 
than  a  quadrant. 

CoR.  2. — If  the  two  sides  are  together  greater  than  a 
semi-circle,  the  interior  angle  will  be  greater  than  the  ex- 
terior and  opposite  angle,  and  the  sum  of  the  angles  at  the 
base,  will  be  greater  than  two  right  angles,  and  the  great- 
er side  will  be  greater  than  a  quadrant.  Conversely,  if 
the  interior  is  greater  than  the  exterior  angle,  then  the  two 
interior  angles  are  together  greater  than  two  right  angles, 
and  the  two  opposite  sides  are  together  greater  than  a  semi- 
circle. 

Thatis,  IfCA-f-CE<AD, 

Then  CAE>CEA,  and  CAE+CEA  >2  right  angles, 
and  conversely,  if  CAE>CED,  then  CAE-f  CEA>  2 
right  angles,  and  CA  +  CE  >  AD. 

Also  if  the  two  sides  be  less  than  a  semi-circle,  the  inte- 
rior angle  will  be  less  than  the  exterior  and  opposite  angle, 
and  the  two  angles  at  the  base,  will  be  less  than  two  right 
angles.     Also  the  less  side  will  be  less  than  a  quadrant. 

That  is.  ifCA-fCE<AD,  then  CAE<CED; 

And  CAE-|-CEA<2  right  angles,  and  conversely. 

Also  the  less  side  is  less  than  a  quadrant. 

CoR.  3. — Hence  if  each  of  the  sides  of  any  spherical 
triangle,  is  greater  than  a  quadrant,  as  any  two  of  them  are 
greater  than  a  serai-circle,  the  sum  of  any  two  angles  is 
greater  than  two  right  angles,  and  therefore  each  of  the  an- 
gles of  the  triangle,  is  greater  than  a  right  angle;  and  con- 
versely. 


144  bPHERICAL  GEOMETRY. 


For  a  similar  reason,  if  each  side  is  less  than  a  quadrant, 
each  angle  will  be  less  than  a  right  angle,  and  conversely. 
And  if  each  side  is  equal  to  a  quadrant,  each  angle  will  b^ 
equal  lo  a  right  angle,  and  conversely. 


PROPOSITION  XIII. 

If  from  the  extremities  of  the  base  of  a  spherical  trian- 
gle arcs  of  great  circles  be  described,  meeting  in  a  point 
within  the  triangle,  these  arcs  shall  together  be  less  than  the 
two  sides  of  the  triangle  ;  and  if  two  sides  of  the  triangle 
are  together  not  greater  than  a  semi-circle,  the  arcs  shall 
make  a  greater  angle,  than  that  made  by  the  sides  of  the 
triangle. 

The  former  part  of  this  proposition  follows  from  Prop. 
VI,  and  the  latter  from  Prop.  XII,  in  the  manner  of  plain 
triangles.     (See  Euclid  1,  21.) 

The  necessity  of  the  condition  in  the  latter  partis  owing 
lo  the  difference  between  spherical  and  plain  triangles  be- 
fore noted.     (See  Prop,  XI,  also  Euclid  1, 16,  2].) 


PROPOSITION  XIV. 

In  any  right  angled  spherical  triangle,  the  sides  are  of 
the  same  affection  as  the  opposite  angles;  that  is,  if  either 
side  is  greater  or  less  than  a  quadrant^  the  opposite  angle  ii 
greater  or  less  than  a  rifrht  angle. 

V\^.  10.  ^^^  ABC,  be  a  spherical  triangle,  right  angled  at  A,  ei- 
ther side  as  AB  will  be  of  the  same  affection  with  the  oppo- 
site angle  A(yB. 

Fig.  12.  ^^^  -^^  ^^  '^ss  than  a  quadrant,  let  AE  be  a  quadrant, 
and  let  EC  be  a  great  circle  passing  through  E,  C.  Since 
A  is  a  right  angle,  and  AE  a  quadrant,  E  is  the  pole  of 
the  great  circle  CA,  and  EC  A  is  a  right  angle.  But 
EGA  is  greater  than  BCA,  therefore  BCA  is  less  than  a 
right  angle. 


SPHERICAL  GEOMETRY.  145 

Let  AB  be  greater  than  a  quadrant,  make  AE  a  quad- 
rant, and  let  a  great  circle  pass  through  C,  E  ;  EGA  is  ^orpf  XV 
right  angle  as  before,  and  BCA  is  greater  than  ECA  that  Fig.  6. 
is  e;reater  than  a  right  angle. 

q.  E.  D. 


PROPOSITOIN  XV. 

If  the  two  sides  of  a  right  angled  spherical  triangle,  beef 
«he  same  affection,  the  hypotenuse  will  be  less  than  a 
quadrant  if  they  be  of  different  affection,  the  hypotenuse 
will  be  greater  than  a  quadrant ;  and  if  each  side  be  equal 
to  a  quadrant,  the  hypotenuse  will  be  a  quadrant. 

Let  ABC  be  a  right  angled  spherical  triangle,  if  the  twoPl-  X^- 
sides  AB,  AC  be  of  the  same  or  different  affection,  the  '^'  ^ 
hypotenuse  BC  will  be  less  or  greater  than  a  quadrant. 

L — Let  AB,  AC  be  each  less  than  a  quadrant.  Let 
AE,  AG  be  quadrants,  G  will  be  the  pole  of  AB,  and 
E  the  pole  of  AG,  and  EC  a  quadrant ;  but  (L  Prop.  IV.) 
CE  is  greater  than  CB,  since  CB  is  farther  off  from  CGD, 
than  CE.  In  the  same  manner,  it  is  shown  that  CB,  in 
the  triangle  CBD,  where  the  two  sides  CD,  BD  are  each 
greater  than  a  quadrant,  is  less  than  CE,  that  is  less  than  a 
quadrant. 

2. — Let  AC   be   less,    and   AB   greater  than    a    quad-  ^''g-  ^^ 
rant,  then  the  hypotenuse  BCwill  be  greater  than  a  quad- 
rant, for  let  AE  be  a  quadrant,  then  E  is  the  pole  of  CA, 
and    EC    is  a    quadrant.     But    CB  is    greater  than   CE 
(\.  Prop.lV.)  since  AC  passes  through  the  pole  of  ABD. 

Q.  E,  D. 

Cor.  1. — Conversely,  accordingly  as  the  hypotenuse, 
is  greater,  less,  or  equal  to  a  quadrant,  the  sides  will  6e  of 
different,  or  of  the  same  affection,  or  will  be  quadrants. 
Also  the  oblique  angles  will  be  of  different  or  of  the  same; 
affection,  or  right  angles.     (XIV.) 


ik^  SPHERICAL  GEOMETRY. 

Con.  2. — When  an  angle  and  the  side  adjacent  are  of 
the  same  affection,  the  hypotenuse  is  less  than  a  quad- 
rant, and  conversely. 

Cor.  3.— If  in  a  ri^ht  angled  spherical  triangle,  one 
side  be  less  than  a  quadrant,  it  is  less  than  the  hypotenuse; 
for  its  opposite  angle,  is  less  than  a  right  angle,  and  there- 
fore the  side  is  less  than  the  hypotenuse  which  is  opposite 
the  right  angle. 

For  a  similar  reason,  if  one  side  be  greater  than  a  quad- 
rant, it  is  greater  than  the  hypotenuse.  And  if  it  be  equal 
to  a  quadrant,  it  is  equal  to  the  hypotenuse. 

CoR.  4. — In  a  right  angled  spherical  triangle,  the  meas- 
ure of  an  acute  angle  is  not  less  than  the  opposite  side  ; 
and  the  measure  of  an  obtuse  angle  is  not  greater  than  the 
opposite  side. 


PROPOSITION   XVI. 

In  any  spherical  triangle,  if  the  perpendicular  from  one 
of  its  angles  upon  the  base,  fall  within  the  triangle,  the  an- 
gles at  the  base  will  be  of  the  same  affection,  but  if  it  fall 
without  the  triangle,  the  angles  will  be  of  different  affection. 

Vis-  ^^-  Let  ABC  be  a  spherical  triangle,  and  letthe  arc  be  drawn 
from  C  perpendicular  to  the  base  AB. 

1.  Let  CD  fall  within  the  triangle  ;  then,  since  ADC, 
BDC  are  right  angled  spherical  triangles,  the  angles  A,  B 
must  each  be  of  the  same  affection  with  CD.     (XIV.) 

>'ig-  1^-  2.  Let  CD  fall  without  the  triangle  ;  then  the  angle  B  is 
of  the  same  affection  with  (.'D;  and  the  angle  CAD  is  of 
the  same  affection  with  CD  ;  therefore  the  angle  CAD 
and  B  are  of  the  same  affection,  and  the  angle  CAB  and 
B  are  therefore  of  different  affections. 

Q.  jE.   D. 

(^OR. — Conversely,  if  the  angles  at  the  base  of  any 
triangle,  be  of  the  same  affection,  the  perpendicular  will 
fall  within  the  triangle,  but  if  the  angles  be  of  different  af- 
fection, the  perpendicular  will  fall  without  the  triangle. 


SPHERICAL  GEOMETRY.  l47 


PROPOSITION  XVII. 


A  perpendicular  being  drawn  to  the  base  of  a  spherical 
triangle,  if  the  sum  of  the  sides  is  less  than  a  semi-circle, 
then  the  least  segment  of  the  base  is  adjacent  to  the  least 
side  of  the  triangle;  but  if  the  sum  of  the  sides  be  greater 
than  a  semi-circle,  the  least  segment  is  adjacent  to  the 
greatest  side. 

Let  ABEF  be  a  great  circle  of  a  sphere,  H  its  pole,  and 
GHDany  circle  passing  through  H,  which  therefore  is  per- 
pendicular to  the  circle  ABEF.  Let  A  and  B  be  two  points 
in  the  circle  ABEF,  on  opposite  sides  of  the  point  D,  and 
let  D  be  nearer  to  A,  than  to  B,  and  let  C  be  any  point  in 
the  circle  GHD  between  H  and  D.  Through  the  points 
A  and  C,  P.  and  C,  let  the  arcs  AG  and  BC  be  drawn, 
and  let  them  be  produced,  till  they  meet  the  circle  CBEF 
in  the  points  E  and  F,  then  the  arcs  ACE,  BCF  are  semi- 
circles. 

Also,  ACB,  ACF,  CFE,  ECB  are  four  spherical  tri- 
angles,  contained  by  arcs  of  the  same  circles,  and  having 
the  same  perpendiculars  CD  and  CG. 

1.  Now  because  CE  is  nearer  to  the  arc  CHG  than  CB 
is,  CE  is  greater  than  CB,  and  therefore  ('E  and  CA 
are  greater  than  CB  and  CA;  wherefore  CB  and  CA  are 
less  than  a  semi-circle;  but  because  AD  is  by  supposition 
less  than  DB,  AC  is  also  less  than  CB.  (I.  Prop.  IV.)  and 
therefore  in  this  case,  viz.  when  the  perpendicular  falls 
within  the  triangle,  and  when  the  sum  of  the  sides  is  less 
than  a  semi-circle,  the  least  segment  is  adjacent  to  the 
leastside. 


2.  Again,  in  the  triangle  FCA  the  two  sides  FC 
and  CA  are  less  than  a  semi-circle;  for  since  AC  is  less 
than  CB,  AC  and  CF  are  less  than  BC  and  OF. 


Fi-.  11. 


148  SPHERICAL  GEOMETRY. 

Also  AC  is  less  than  CF,  because  it  is  more  remote 
from  CHGthan  CF  is,  therefore  in  this  case,  also. viz,  when 
the  perpendicular  falls  without  the  triangle,  and  when  the 
sum  of  their  sides  is  less  than  a  semi-circle,  the  least  seg- 
ment of  the  base  AD  is  adjacent  to  the  least  side. 

3.  But  in  the  triangle  FCE,  the  two  sides  FC  and  CE 
are  greater  than  a  semi-circle;  for  since  FC  is  greater  than 
CA,  FC  and  CE,  are  greater  than  AC  and  CE.  And 
because  AC  is  less  than  CB,  EC  is  greater  than  CF,  and 
EC  is  therefore  nearer  to  the  perpendicular  CHG  than  CF 
is,  wherefore  EG  is  the  least  segment  of  the  base,  and  is 
adjacent  to  the  greatest  side. 

4.  In  the  triangle  ECB,  the  two  sides  EC,  CB  are  great- 
er than  a  semi-circle,  for  since  by  supposition  CB  is  threat- 
greater  than  CA,  EC  and  BC  are  greater  than  EC  and 
CA. 

Also,  EC  is  greater  than  CB;  wherefore  in  this  case 
also,  the  least  segment  of  the  base  FG  is  adjacent  to  the 
greatest  side  of  the  triangle.  Wherefore,  when  the  sum  of 
the  sides  is  greater  than  a  semi-circle,  the  least  segment  of 
the  base  is  adjacent  to  the  greatest  side,  whether  the  per- 
pendicular fall  within  or  without  the  triangle  :  and  it  has 
been  shewn,  that  when  the  sum  of  the  sides  is  less  than  a 
semi-circle,  the  least  segment  of  the  base  is  adjacent  to 
the  least  of  the  sides,  whether  the  perpendicular  fall 
within  or  without  the  triangle.* 


Q.  E,  D. 


PROPOSITION  XVIII. 


The  suras  of  the  opposite  angles  of  a  quadrilateral  figure, 
PI    xvi.^"  ^^®  sphere,  which  can  be  inscribed  in  a  small  circle,  are 

equal 
Fig.  7         That  is  A+C=B+D. 


*  When  the  perpendiculars  fall  without  the  triangle,   it   is  that  whirh 
is  noarest  to  the  triftr.^le  to  which  the  demonstration  is  applicable. 


SPHERICAL  GEOMETRY.  140 

For  if  arcs  of  great  circles  pass  through  the  Pole  of  the 
small  circle  and  each  angle  of  the  quadrilateral,  they  will 
divide  it  into  isosceles  triangles,  (Def.  9.  Cor  7.)  whose  an- 
gles at  the  base  in  each  will  be  equal,  (IV.)  and  conse- 
quently the  sums  of  equals  will  be  equal. 

That  is  PBA+PBC-f  PDA+PDC=PAB4-PAD+ 
PCD-fPCB. 

Or  ABC  +  BDC=BAD4-BCD. 


q.  E.  D. 


PROPOSITION  XIX. 


If  two  spherical  triangles,  have  two  right  angles,  in 
other  words,  If  their  sides  intersect  each  other  in  the  Pole 
of  their  base,  then  the  two  triangles  are  to  each  other  as 
their  bases.  p.^ 

If  a  great  circle  should  pass  through  P,  and  bisect  the  ^*'  " 
base,  EG,  of  the  triangi-  PEG,  it  would  also  bisect  the 
triangle,  (  V  )  for  the  two  triangles  into  which  it  would 
divide  PEG  would  be  equal.  Also  if  any  multiple  of  the 
base  should  be  taken,  the  triangle  formed  by  a  great  cir- 
cle passing  through  its  extremity  and  the  Pole  (P  )  would 
be  the  same  multiple  of  the  triangle  PEG. 

Therefore  the  demonstration  of  Euclid  (VI,  33.)  con- 
cerning the  angles  at  the  centre  of  a  circle,  and  the  circu- 
lar sectors,  is  applicable  in  every  respect,  to  the  angles 
at  the  Pole  of  a  great  circle  on  the  sphere,  and  to  the  tri- 
angles formed  between  it,  as  a  vertex,  and  the  arcs  of  that 
circle  as  bases. 

That  is,  the  angles  at  the  Pole,  and  also  the  triangles 
themselves,  have  to  each  other  respectively,  the  same  ra- 
tio, which  their  bases  have. 

q.  E.  D. 

Cor.  1 . — In  such  triangles,  the  area,  varies  as  the  base, 
or  as  the  angle  at  the  Pole,  of  which  the  base  is  a  meas- 
ure ;  that  is  the  base  is  a  measure  of  the  area  of  the  trian- 
gle. 

CoR.  2. — Hence,  since  the  triangle  E;7G,  equals  EPG 
in  every  respect,  therefore,  the  arc  EG  which  is  the  meas- 

20 


150  SPHERICAL  GEOMETRY. 

ure  of  the  angle  at  the  Pole,  is  also  the  measure  of  the 
lune  FEpG  ;  which  is  composed  of  the  two  triangles. 

Thus,  when  the  lune  is  right  angled,  or  EG=  a  quadrant, 
it  is  one  half  of  a  hemisphere,  and  as  EG  approaches  to 
a  semi  circle,  the  lune  approaches  to  a  Hemisphere. 

Also  a  spherical  triangle  which  has  its  three  angles 
right,  being  the  half  of  a  right-angled  lune,  is  equal  to  one 
fourth  of  a  Hemisphere. 


PROPOSITION  XX. 

As  Four  right  angles,  is  to  the  excess  of  the  angles  of 
any  spherical  triangle  above  two  right  angles ;  called  the 
Spherical  Excess  ;  so  is  the  area  of  the  Hemisphere,  to  the 
area  of  the  triangle. 

Let  ABC  be  a  spherical  triangle,  BCEF,  be  the  cir- 
Pl.  XVI.  cle,  of  which  BC  is  an  arc  ;  let  ABDCA,  be  the  lune  for- 
^»g-  6.  med  by  the  two  semi-circles,  of  which  AB  and  AC  are 
arcs,  and  the  parts  CAF,  BAE  two  semi-circles,  meeting 
the  circle  BCEF,  (I,Prop.  I  Cor.  1 0,)  then  the  triangle  EFA 
=BDC  on  the  opposite  Hemisphere  (II)  therefore  the  lune 
ABDC,  which  equals  ABC+DBC,  equalsalso  ABC-f 
AFE. 

Also  ABC-f  ABF=  lune  CAFB, 

and    ABC4-ACE=luneBAEC, 

and  ABC  +  AFE  +  ABF-f  ACE=  the  Hemisphere. 

Now  these  lunes  are  to  each  other  respectively  as  their 
angles  (XIX,  Cor  2.),  that  is,  as  the  angles  of  the  spher- 
ical triangle,  and  each  of  them,  is  to  the  Hemisphere  as 
its  angle,  is  to  two  right  angles, 
That  is,  2.  right  angles  :  angle  A:  :Hem.  :  ABC-j-AFE, 
and       2.     „       „       :      „     B::Hem.  :  ABC  +  ACE, 
2.    „      „      :      „     C: :  Hem.  :  ABC-f  ABF, 
two  right  angles  are  to  the  angles  at  A,  B,C,  taken 
5. 12.  together,  as  a  Hemisphere  is  to    3    ABC+AJE  +  ACE-f 
ABF  ;  therefore  also  two  right  angles,  are  to   the   Excess 
and^D  ^^  A-fB-f-C,  above  two  right  angles,  as  a  Hemisphere  is 


SPHERICAL  GEOMETRY.  151 

to  the  excess   of   3ABC-f  AFE-f-ACE-fABF,  above  a 
Hemisphere,  that  is  to  2  ABC. 

Or  4  right  angles  :  Spherical  Excess  .''IS  Hera.  :  2ABC, 

::    Hem.:    ABC. 
Q.  E.  D. 

Cou.  -  As  the  first  and  third  terms  are  constant,  there- 
fore, by  equality  of  ratios  the  Spherical  Excess  is  the 
measure  of  the  area,  of  a  spherical  triangle. 


III.  INTERSECTIONS 

IN  THE 
PLANE  OF  A  GREAT  CIRCLE  OF  THE  SPHERE. 

PROPOSITION  L 

If  a  line  be  drawn  from  any  point  of  the  Sphere,  to  the 
pole  of  a  great  circle,  the  distance  from  the  centre  of  that 
circle,  to  the  point  where  the  line  meets  its  plane,  is  equal 
to  the  tangent  *  of  half  the  distance  of  the  given  point,  from 
the  other  pole  of  the  circle. 
Fig-     18.     That  is  Cf=pi,  where  /?0=|joE. 

For  the  triangles  CP/=/)C^,  are  similar  and  equal  in  all 
respects. 

C-  JS.  D. 


PROPOSITION  II. 

If  one  extremity  of  a  straight  line,  remain  in  the  pole  of 
a  great  circle,  while  the  line  is  carried  around  in  the  cir- 
cumference of  any  parallel  circle,  it  will  describe  the  con- 
vex surface  of  a  right  cone ;  and  the  intersection  of  this 
surface,  with  the  plane  of  the  great  circle,  is  a  circle  whose 
centre  will  coincide  with  that  of  the  Primitive  circle. t 

*  The  word  tangent  is  here  used,  a3  in  Trigonometry,  for  that  part  of  a 
line  touching  a  circle,  which  is  intercepted  between  the  point  of  contact, 
and  any  diameter  of  the  circle  produced. 

Also,  distance  on  the  sphere,  is  used  for  the  arc  of  a  great  circle^  which 
passes  through  any  two  points  ;  for  this  arc  on  the  sphere,  like  a  straight 
line,  in  a  plane,  is  the  least  line  which  can  be  drawn  between  two  points. 

t  The  term  Primitive  is  here  applied  to  that  great  circle,  in  whose 
plane  the  intersections  are  supposed  to  be  made. 


SPHERICAL  GEOMETRY.  I53 

1.  The  Figure  FPI  described  by  the  motion  of  the  line,  fig.  17, 
is  evidently  a  right  cone.  (See  EucHd,  Sup.  Ill,  def.  11.) 

2.  Its  intersection  with  the  plane  of  the  Primitive  is  a 
circle.     (Con.  Sect.  def.  4.) 

3.  The  centre  will  coincide  with  that  of  the  Primitive, 
for  the  axis  of  the  cone,  PH,  is  the  axis  of  the  sphere  and 
of  the  great  circle  EQ,  through  the  centre  of  which,  and 
of  all  circles  parallel  to  it,  it  passes.     (Def.  9,  Cor.  3.) 

Q.  E.  D. 

Cor. — If  the  vertex  of  the  cane,  or  P,  be  the  more  dis- 
tant pole,  the  circular  intersection  will  be  within  the 
Primitive,  as  Jhi,  and  consequently  less  than  the  Primi- 
tive. But  if  the  vertex  be  the  nearer  pole,  the  intersec- 
tion will  be  without  the  Primitive,  and  larger  asft* 


PROPOSITION  HI. 

The  same  being  supposed  as  in  the  last  proposition,  but 
the  circle  oblique  to  the  Primitive,  the  figure  described 
will  be  an  oblique  cone,  and  its  intersection  with  the  Prim- 
itive will, be  a  circle. 

1 .  The  figure  described  FPI  is  an  oblique  cone,     (Con.  Fig,  le. 
Sect.  Appendix  Def.  1.) 

2.  Its  intersection  with  the  Primitive,  fhi^  is  a  circle,  i-  29, 
for  it.is  a  sub-contrary  section,  since  IPx  =11/,  and  also=c  ^'  ^^* 
IFF.  (Con.  Sec.  App.  VI.) 

Q.  E,  D. 

Con.  1 . — If  F  be  between  E  and  P,  a  part  of  the  in- 
tersection will  be  in  the  plane  of  the  Primitive,  and  the  re- 
mainder without  it,  and  if  the  oblique  circle  is  a  Great 
Circle,  the  corresponding  circular  intersection,  will  cut  the 
primitive  in  two  points  diametrically  opposite.  (Fugf.  I. 
Cor.  10.) 

The  preceding  proposition  and  demonstration,  are  ap- 
plicable to  any  small  circle  which  is  at  right  angles  to  the 


154  SPHERICAL  GEOMETRY. 

Fig^  19.  primitive  as  EFI,  whose  corresponding  circular  intersec- 
tion isfhi. 


PROPOSITION  IV. 

The  centres  of  the  circular  intersections  in  the  plane  of 
the  Primitive  arc,  in  all  cases,  are  in  the  line  of  mutual  in- 
tersection of  the  Primitive  and  a  plane  passing  throug:h  the 
centre  of  the  sphere,  and  through  the  poles  of  the  Primi- 
tive, and  of  the  given  circle,  upon  its  surface. 
Fig,  16.  That  is,  the  centre  oi  fhi  ism  QE,  which  is  the  inter- 
17.  &  l9gection  of  the  Primitive,  and  of  I^E,  or  a  great  circle  pass- 
ing through  the  poles  of  the  Primitive,  and  of  the  circle 
FHI. 

For  the  circle  I/^DE,  is  perpenrl'cular  to  the  Primitive, 
and  to  the  circle  FHI,  {Dei,  9.  Cor.  4.)  and  coincides  with 
the  axes  of  both,  and  consequently  passes  through  the  centres 
of  both,  bisecting  both  thecircle  FHI,  on  the  sphere,  and  the 
corresponding  circularintersectjon  in  the  plane  of  the  prim- 
itive. 

Q.  £.  D. 

Cor.  1. — If  the  circle  on  the  sphere  be  an  oblique  great 
circle,   the  centre   of  the  corresponding  circular  intersec- 
tion will  be  in   that  diameter  of  the  Primitive  produced, 
which  is  perpendicular  to  its  intersection  with  the  oblique 
Fig.  23.  circle. 
3.  1.        That  is,  in  QEt,  which  is  perpendicular  to  FI. 

CoR.  2. — Hence  if  any  number  of  oblique  great  circles, 
cut  the  Primitive  in  the  points  F  and  I,  the  centres  of  all 
their  circular  intersections,  will  be  in  the  same  line.  Viz. 
in  QEt  produced. 

Fig.  20.  CoR.  3. — In  like  manner  if  a  small  circle  be  at  right 
angles  at  the  Primitive,  the  centre  of  the  corresponding 
circular  intersection,  will  be  that  diameter  of  the  Primitive, 
which  is  perpendicular  to  its  intersection  with  the  small 
circle. 

And  the  centres  of  all  such  circles,  will  be  in  the  same 
straight  line,  Ei  produced. 


SPHERICAL  GEOMETRY.  155 

Cor.  4. — If  any  number  of  great  circles  oblique  to  the  ^»g-  21. 
Primitive,  intersect  each  other  m  one  point  on  the  Sphere, 
their  corresponding  circular  sections,  will  intersect  each  oth- 
er in  one  point  in  the  plane  of  the  Primitive,  and  their  cen- 
tres will  be  in  a  straight  line  KL,  perpendicular  to  the  di- 
ameter of  the  Primitive  which  passes  through  the  point  of 
their  mutual  intersection,  and  passes  through,  h,  the  centre 
of  the  circular  lection  whose  chord,  at  its  intersections  with 
t-he  Primitive,  is  parallel  to  KL. 


PROPOSITION  V. 

The  distance  between  the  centres  of  the  Primitive  and  of 
any  circular  intersection  of  a  great  circle  in  its  plane,  is 
equal  to  the  tangent  of  the  arc  which  measures  the  inclina- 
tion of  the  circles  on  the  sphere,  I.  Prop.  Cor.  I. 

That  isPA=Q^  Eig.  23. 

For  APQ,  being  the  inclination  of  the  circles  on  the 
sphere,  and  Q^,  the  tangent  of  QA,  which  measures  their 
inclination. 

Then  QPA=:FPK;  h  FPK-hFPA=right  angle.         ^. 
Also     AIP+PU4-HIi*=PlA-fPAI  =  right  angle,  ^\\'^ 

.:.     AlF'hhli=Phl=h[i+hil=2hli; 

.:.     A\P=hli ;  h  2AIP,  or  APF=2hIz=PM, 

.:.     APF-fPIA=FPK4-APF, 

.-.     PIA=FPK=APQ^  hPh=qt. 

Q.  E.  D, 

CoR.  1. — Hence  \h  the  radius  of  the  circular  intersec- 
tion F/Ii,  is  equal  to  P^,  the  secant  of  the  arc  QA,  which 
measures  the  inclination  of  the  cicles. 

CoR.  2. — PA:,  the  distance  from  the  centre  of  the  Prim- 
itive to  the  point,  where  a  line  connecting  the  pole  of  the 
oblique  circle  with  that  of  the  Primitive,  meets  the  plane  of 
the  latter,  is  equal  to  the  tangent  of  half  the  the  angle  of 
their  inclination. 

F0rPIA:=iFPK=iAPQ. 


n^ 


SPHERICAL  GEOMETRY 


PROPOSITION  VI. 


If  a  small  circle  be  at  right   angles  to  the  Primitive,  the 
radius  of  its  circular  intersection,  will  be  the  tangent  of  its 
distance  from  the  nearer  pole. 
Fig.  20.      That  is,  Fh  is  the  tangent  of  FE. 

For  xDF=DBF=PFB, 

And  x'Di=hiF=hFi, 
3.  31.         .-.  PFB=AFi,  .-.  PF/i-:/Fi=right  angle. 
.'.     Fh  is  the  tangent  of  FE. 

Cor. — Fh  the  distance  of  the  centre  of  the  circle  F/  li, 
from  that  of  the  Primitive,  is  the  Secant  of  FE,  the  distance 
of  the  small  circle  from  its  nearest  pole. 


PROPOSITION  VII. 

The  angle  made    by  two  circles    which  cut  each   oth- 
er on  the  sphere,  is  equal  to  the  angle  made  by  the  corres- 
ponding circular  intersections  in  the  plane  of  the  Primtive. 
Fig.  22.      That  is,  DFK,ar  D  FK'=ci/A;  or  DfK', 

For  suppose  planes  to  coincide  with  FK  and  FD,  the 
tangents  to  the  circles,  at  the  point  of  their  intersection, 
and  to  pass  through  P ;  these  planes  will  touch  the  con- 
vex surfaces  of  the  cones,  whose  bases  are  the  two  circles. 
Also  let  another  plane  pass  through  P,  and  cut  those  tan- 
gents, in  K  and  D,  forming  a  pyramid  PKDG,  which  is 
cut  by  a  plane  KDG,  parallel  to  the  primitive.  Let  FI  be 
parallel  to  EQ. 
^  32.        Then  PIF=PFK'  =  GFK,  also  PIF=PFI=FGK, 

Therefore  FK=KG. 

In  like  manner  FD=DG. 


SPHERICAL  GEOMETRY. 


m 


Therefore  KFD=KGD=^/J,  in  a  plane  parallel  to  ^  '^^ 
KGD.  But  since  the  planes  PFK,  PFD  touch  the  cones, 
their  intersection  with  the  primitive,  will  be  straight  lines 
touching  the  circular  intersections,  in  the  same  plane. 
And  the  angle  made  by  two  circles  which  cut  each  other, 
in  a  plane,  is  here  considered  as  equal,  or  is  the  same  as 
the  angles  made  by  lines  touching  them  at  the  point  of 
intersection. 

In  most  cases,  the  demonstration  will  be  much  more  con- 
cise and  simple,  by  taking  the  vertical  angle  of  the  tangents, 
viz:  D'FK',  and  proving  its  equality  with  D/ K'  in  the 
plane  of  the  Primitive. 

For  DTP=FIP=DyF,  therefore  D'F=Dy. 

In  like  manner  KT=K/,  and  therefore,  D  FK'  D'/K'. 

Q.  E.  b. 


PROPOSITION  VIII. 

If  on  the  sphere,  any  number  of  small  circles  pass  through 
one  pole  of  the  Primitive,  and  the  farther  pole  of  any  ob- 
lique great  circle,  their  circular  intersections  will  all  pass 
through  the  centre  of  the  Primitive,  and  also  through  the 
point,  where  a  line  drawn  from  the  other  pole  of  the  Prim- 
itive through  the  pole  of  the  oblique  circle,  meets  the  plancp^''  ^j.^' 
of  the  Primitive. 

That  is,  the  circular   intersections,  will  all  pass  through 
P  andy. 

This  is  evident  from  the  demonstrations  of  the  preceding 
propositions. 


PI.  XIV. 


Cor.  1. — MO,  ON  are  respectively  equal  to  the  arcs 
wo,  on,  of  the  oblique  circle  on  the  sphere. 

CoR.  2. — ^The  intersections  of  the  planes  of  the  small  Fig.  4. 
circles,  with  the  Primitive,  are  straight  lines  which  intersect 
each  other,  in  the  point  (y)  where  aline  drawn  from  the 
given  pole  of  the  oblique  circle,  to  that  of  the  Primitive, 
meets  the  plans  of  the  latter. 

21 


tSB  SPHERICAL  GEOMETRY. 

Pj-  XL        PB,  PN,  PX,  are  the  intersections  of  the  pimitive,  and  of 
*^*     '  the  planes  of  small  circles,  passing  through  its  pole,  and  the 
farther  pole  of  the  oblique  circle  CRSOD. 

Cor.  3.— RS  and  SO  correspond  to  arcs  of  the  oblique 
circle  on  the  sphere,  respectively  equal  to  XN  and  NB. 


ivr 


APPENDIX  TO  SPHERICAL  GEOMETRY 


13 


PROJECTIONS. 


Def.  1. — To  Project  oin  object^  is  to  represent  every 
point  of  it,  on  a  plane,  as  it  appears  to  the  eye  in  a  certain 
position. 

2. — The  Plane  of  Projection,  is  that  on  which  the  object 
represented. 

3. — The  Projecting  Pointy  is  that  point  where  the  eye 
is  supposed  to  be  placed. 

4, —The  Orthographic  Piojection  of  the  sphere  is  thai 
in  which  a  great  circle  is  assumed  as  the  plane  of  projection, 
and  a  point  at  an  infinite  distance  in  the  axis  produced,  as 
the  projecting  point. 

5. — The  Stereographic  Projection  of  the  sphere  is  thai, 
in  which  a  great  circle  is  assumed  as  the  plane  of  projec- 
tion, and  one  of  its  poles,  as  the  projecting  point. 

6. — The  Gnomonic  Projection  of  the  sphere,  is  that  in 
which  the  plane  of  projection  touches  the  sphere,  and  the 
centre  is  the  projecting  point. 

ScHOL. — In  the  Theory  of  Projections,  the  rays  of  light 
are  supposed  to  move  in  strai^iht  lines,  from  any  point  of  an 
object  to  the  eye.  and  the  projection  of  each  point,  is  the 
intersection  of  such  a  line,  and  the  plane  of  projection. 

7. — A  direct  circle  is  parallel  to  the  plane  of  projection. 


160  APPENDIX  TO  SPHERICAL  GEOMETRY. 

8. — ^n  oblique  circle  is  oblique  to  the  plane  of  projec- 
tion. 

9. — M  right  circle,  is  that  whose  plane  coincides  with 
the  axis  of  the  eye. 

Cor. — A  Great  Circle  which  is  right,  is  perpendicular 
to  the  plane  of  projection. 

The  following  results  or  Laws  of  Projection  will  follow 
as  Corollaries,  from  the  preceding  definitions  and  Scholi- 
um. 


I.  Of  Orthographic  Projection »  N 

.)( 

1. — The  Rays  of  light  being  supposed  to  come  from  an 
indefinite  distance,  'ttiay  be  considered  as  parallel  to  each 
oilier,  and  perptndicular  to  the  plane  of  projection, 

2. — A  straight  line,  perpendicular  to  the  axis  of  the  eye, 
is  projected  into  a  point. 

3. — A  straight  line  parallel  to  the  plane  of  Projection,  is 
projected  into  a  line  equal  to  itself. 

4. — A  straight  line  oblique  to  the  plane  of  Projection  is 
projected  into  a  line  less  than  itself,  in  the  ratio  of  the  sine 
of  the  angle  which  it  makes  with  any  ray  of  light  to  radius, 

6. — So  a  plane,  perpendicular  to  the  plane  of  Projection, 
is  projected  into  a  straight  line,  and  a  plane  parallel  to  the 
plane  of  Projection,  is  projected  into  a  plane  equal  and 
similar  to  itself. 

n.  VIII.       7. — A  circle  oblique  to   the  plane  of  Projection  is  pro- 
='  ^  '  jected  into  an  Ellipse. 

For  let  AMB  represent  the  circle  to  be  projected,  and 
ALB  the  figure  into  which  it  is  projected.  Any  ordinates 
in  the  circle  will  be  projected  into  lines  less  than  them- 
sel^fcs,  in  the  same  ratio.     (4.) 


APPENDIX  TO  SPHERICAL  GEOMETRY.  161 

Thatis,KM:DG::KL:DE; 

which  is  the  property  of  an  Ellipse. 

As  Orthographic  Projection,  is  not  necessarily  hmited  m 
its  application,  to  the  sphere,  having  no  particular  relation 
to  it,  and  is  principally  used  in  a  noore  general  application  to 
mathematical  figures,  it  is  sufficient  here,  to  give  a  mere 
sketch  of  its  general  properties. 

11.  Stereographic  Projection,  is  confined  to  the  sphere, 
and  on  many  accounts,  is  the  most  convenient  method  of 
representing  on  a  plane,  figures  on  the  surface  of  the  sphere. 

By  comparing  the  preceding  definitions  with  the  propo- 
sitions, in  the  third  part  of  Spherical  Geometry,  it  will  be 
evident, 

1. — That  all  circles,  on  the  surface  of  the  sphere,  are 
projected  into  straight  lines,  or  into  circles.  This  renders 
the  practical  operations  in  this  method  of  projection  very 
easy. 

2. — That  these  projected  circles,  in  all  cases,  make  the 
same  angle,  on  the  plane  of  Projectioii,  which  the  circles 
to  be  projected  make  on  the  sphere. 

3. — That  the  centres  of  projected  circles  and  their  pro- 
jected poles  are  accurately  and  easily  found,  by  geometri- 
cal operations. 

III.  Gnomonic  Projection,  has  a  particular  reference  to  the 
sphere,  but  is  of  a  very  limited  application  ;  being  used  to 
explain  the  theory  of  Dialing,  or  the  Geometrical  construc- 
tion of  Dials. 

If  the  Earth  were  supposed  transparent,  and  its  axis  ca- 
pable ofcasting  a  shadow,  this  shadow  from  the  Earth's  ro- 
tation would  coincide  successively  with  the  planes  of  dif- 
ferent Meridians  ;  and  as  the  rotation  of  the  Earth  is  sup- 
posed to  be  uniform,  the  shadow,  would  in  equal  times,  co- 
incide with  Meridians  at  equal  distances  from  each  other, 
or  which  make  Vvith  each  other  equal  angles. 


162  APPENDIX  TO  SPHERICAL  GEOMETRY. 

If  successive  meridians  were  taken,  making  angles  of  15^ 
with  each  other,  the  shadow  of  the  axis  would  move  from 
one  to  the  other  is  one  hour,  or  j\  part  of  an  entire  revolu- 
tion.    Such  Meridians  are  called  hour  circles. 


If  the  planes  of  these  Meridians  or  hour  circles,  weve 
produced  to  intersect  a  plane  touching  the  sphere,  their 
mutual  intersections  would  be  straight  lines,  which  to  an 
eye  placed  at  the  centre  of  the  sphere,  would  coincide  with 
the  planes  of  the  Meridians.  They  would  therefore  be 
the  Projections  of  those  planes,  or  of  their  circumjerences 
on  the  surface  of  the  Sphere.  Hence  in  every  kind  of  Di- 
al, the  edge  or  part  of  the  Gnomon^  which  casts  the  shad- 
ow must  be  placed  parallel  to,  or  coinciding  with  the 
Earth's  axis,  and  supposing  a  sphere  to  surround  it,  the 
hour  lines  are  the  intersections  of  the  hour  circles  with  the 
dial  plate,  which  is  the  plane  of  projection. 

PI.  XV.  Thus  in  the  common  Horizontal  Dial,  CAB,  is  a  triangu- 
p-^  1.  ^^^  plate,  of  which  the  edge,  which  casts  the  shadow,  is 
CA,  and  the  angle  CaB  being  made  equal  to  the  latitude 
of  the  place,  i.  e.  to  the  elevation  of  the  pole,  CA  will  be 
parallel  to  the  axis  of  the  earth.  Suppose  a  sphere  to  sur- 
round CA,  as  its  axis,  whose  centre  is  D,  then  DF  per- 
pendicular to  CA,  will  be  the  intersection  of  its  equator 
with  the  plane  CAB,  andHFG  will  be  the  intersection  with 
the  Dial  plate. 


Making  Fl  and^  each  equal  to  FD,  the  quadrants,  FK, 
fk  will  represent  half  the  Equator,  or  a  circle  concentric 
with  it,  equally  divided  by  Im,  \m  he.  m,  in,  &c.  which 
represent  meridians,  or  hour  circles;  Ym,  F/w,,  &c.  also//i, 
fn,  he.  are  the  intersections  of  these  circles,  with  the  line 
HG,  in  the  plane  of  the  Dial. 

Hour  lines  drawn  from  A  and  a,  through  m,  m,  he.  n, 
n,hc.  will  intersect  the  circumference  of  the  Dial  plate,  in 
the  divisions  which  mark  the  hours. 

The  same  principles,  will  serve  for  the  construction  and 
explanation  of  all  Dials. 


APPENDIX  TO  SPHERICAL  GEOMETRY- 
PROBLEMS 


IN 


STEREOGRAPHIC  PROJECTIO]^. 


PROBLEM  L 

To  project  a  great  Circle,  of  zchich  a  given  point  is  thr 
projected  Pole* 

1 .    TTie  given  point  being  the  centre  of  the  Primitive. 

Then  the  Primitive  itself  is  the  circle  required. 

2,    The  given  point  being  in  the  circumference  of  the  Primi- 

tive. 

Let  the  given  point  be  A.     Draw  through  the  centre  of  Fig.  i 
the  Primitive  the  diameter  ACE,  and  a  perpendicular  di- 
ameter BCD,  which  is  the  projected  circle,  whose  poles  are 
A  and  E. 

Also  B  and  D  are  the  poles  of  the  projected  circle  ACE. 

3.   TTie  given  point,  being  in  the  plane  of  the  Primitive, 

Let  the  given  point  be  p.     Draw  the  diameter  BpD,  i-'i?-  - 
and  its  perpendicular  AC.     From  A  reduce^  to  G,  in  the 
circumference  of  the  Primitive  ;  make  GF  equal  to  a  quad- 
rant, project  F  in  E ;  a  circle  through  A,  E  and  C,  is  the 
projected  circle,  whose  pole  is  p. 


164  appelndix  to  spherical  geometry. 

Cor. — By  reversing  the  process  in  each  case,  the  pro 
jected  pole  of  a  given  projected  circle  may  be  found. 


PROBLEM  IL 

To  Project  a  Small  Circle,  parallel  to  a  given  projected 
Great  Circle,  and  at  a  given  distance  Jrom  it, 

1.  The  given  Great  Circle  being  the  Pnmitive* 

Fig.  15.  Draw  any  diameters  as  BCD,  and  its  perpendicular. 
From  B  set  the  given  distance  to  O;  project  O  in  X,  then 
the  circle  XFE,  is  the  circle  required. 

2.    The  given  Circle  being  a  right  Circle, 

Fig.  16.  Let  the  given  projected  circle,  be  BD.  Set  the  given 
distance  from  B  to  F,  project  F  in  G  on  the  perpendic- 
ular diameter  AY  ;  a  tangent  to  the  primitive,  at  F,  will 
intersect  AY  produced  in  (K)  the  centre  of  the  required 
circle,  which  will  pass  through  F  and  G. 

3.    The  given  Circle  being  oblique. 

Fig.  n.  Let  the  given  oblique  circle,  be  AED.  Find  its  Pole 
(Prob.  L  Cor.)  P  ;  reduce  P  to  N,  and  from  N  set  off  on 
each  side,  the  complement  of  the  given  distance  of  the 
small  circle,  to  R  and  Q ;  project  R  and  Q  in  X  and  Y, 
XY  is  the  diameter  of  the  small  circle  (BXY)  required. 

CoR. — By  reversing  the  process,  the  pole  of  a  given 
small  circle,  parallel  to  a  large  circle,  may  be  found,  anij 
also  the  distance  of  the  small   circle  from  its  large  circle. 


APPENDIX  TO  SPHERICAL  GEOMETRY.  1^5    ** 


PROBLEM  III. 

To  project  a  great  circle^  through  two  given  points  in  the 
Primitive, 


I .  One  of  the  points  being  in  the  centre  of  the  Primitive, 

Let  the  given  points  be  C  and  A,  or  Cand  R.     Then  Fig. 
the  diameter  ACE  or  YRB  will  be  the  circle  required. 


2«   One  of  the  points  being  in  the  circumference  of  the  Primi- 
tive, 

Let  the  given  points  be  C  and  K.     Draw  the  diameter  F'?-  8« 
CXD,  the  circle  required  will  pass  through  C,  K  and  D. 


3.  Keither  of  the  points  being  in  the  centre  or  circumference 
of  the  Primitive* 

Let  the  given  points  be  P  and  Y.     Draw  the  diameter  Fig.  9. 
BPA  and  its  perpendicular  CD,   reduce  P  to   G  ;  the   re- 
quired circle  will  pass  through,  YP  and  G. 


PROBLEM  IV. 

Through  any  given  point,  to  project  two  circles  luhich 
shall  make  a  given  angle. 


1.   When  the  angle  is  at  the  centre  of  the  Primitive. 

Draw  any  diameter  as  ACE,  set  the  measure  of  the  giv-  Fig.  1, 
en  angle  from  A  to  H,  draw  HC,  then  AC  and  HC  are  pro- 
jected circles  making  the  required  angle  at  C. 

22 


166  APPENDIX  TO  SPHERICAL  GEOMETRY. 


2.   When  the  angular  point  is  in  the  circumference  of  the- 
Primitive. 

fiS'  3.  Let  A  be  the  angular  point.  Draw  tbe  diameter  AC, 
and  the  perpendicular  DB.  Set  the  measure  of  the  given 
angle  from  D  to  E;  project  E  to  F,  AFC,  is  a  projected 
circle  which  makes  the  given  angle  with  the  Primitive  at  A. 


3.   When  the  angular  point  is  in  the  plane  of  the  Primitive, 

Fig-  5.  Let  O  be  the  given  point.  Draw  the  perpendicular  di- 
ameters, aOB  and  (JD.  From  C  reduce  O  to  I  •  make 
DK  equal  to  Al ;  project  K  in  P,  CPD  is  the  projected 
circle,  of  which  O  is  the  projected  pole.  (Prob.  1.)  Set 
the  2;iven  angle  from  C  to  L,  let  a  line  from  L  to  O  (which 
will  be  the  projection  of  a  small  right  circle)  intersect  the 
circle  CPD  in  Q,  from  Q  as  a  pole,  project  the  great  circle 
EOF,  (Pkob.  I.)  which  makes  the  given  angle  EOA,  with 
the  right  circle  AOB. 

4.    When  the  circles  required  are  both  oblique. 

Fig.  14.  Let  E  be  the  given  point.  Through  E  project  any  ob- 
lique sreat  circle,  as  AED.  Find  its  pole-  P.  Let  the 
straight  line  EP  intersect  the  Primitive  in  1';  set  off  TV 
the  measure  of  the  given  angle  ;  draw  the  diameter  VY, 
and  its  perpendicular  FN,  NEA,  is  the  required  angle, 
NEF,  -iED,  the  required  circles. 

Cor. — By  reversing  the  process,  in  each  case  the  angle 
made  by  two  given  projected  great  circles  can  be  measur- 
ed. 

PROBLEM  V. 


Through  a  given  point  to  project  a  great  circle  perpen- 
rlirylar  to  a  given  great  circle. 


APPENDIX  TO  SPHERICAL  GEOMETRY.  1^7 

1.    When  the  given  circle  is  the  Primitive. 

Let  Y  or  X  be  the  given  point.     The  diameter  YCB,  or         ^ 
AXCE,  will  be  the  circle  required.  '^" 


2.    When  the  given  circle  is  righty  and  the  given  point  the 
centre  of  the  Primitive* 

Let  AB  be  the  given  projected  circle,  the  perpendicular  ^'^s-  8- 
CXD  is  the  circle  required. 


3.    IVhen  the  given  circle  is  right.,  and  the  given  point  not  in 
the  centre  of  the  Primitive, 

Let  the  given  projected  circle  be  AB,  and  K,  in  it,  the 
given  point.  Draw  the  perpendicular  diameter  CD,  then 
the  required  circle  will  pass  through  C,  K,  and  D. 


4.    When  the  given  circle  is  oblique,  and  the  given  point  in 
the  middle. 

Let  DCK  be  the  given   oblique  circle,  and   K  the  given  Eig.  8. 
point.     The  diameter  AKB,  is  the  required  projected  cir- 
cle. 


5.   When  the  giveyi  circle  is  oblique,  and  the  given  point  not 
in  the  middle  of  it. 

Let  Y  be  the  given  point  in   CYD.     Find  the  poles  P,  f^'g-  ^■ 
p  of  CYD,  and  FYp,  is  the  circle  required. 


PROBLEM  VI. 


To  project  a  great  circle,  through  a  given  point   which 
sjiali  make  a  given  angle  roith  the  Primitive, 

Let  O  be  the  given  point.     Describe  the   circle  CPD,  ^^o'  5. 
whose  pole  is  O,  as  in  Prob.  IV.  3.     Make  BL  equal  to 


168  APPENDIX  TO  SPHERICAL  GEOMETRY. 

th6  given  angle,  project  L  in  M ;  and  describe  the  small 
circle  EQN,  which  is  parallel  to  the  Primitive  and  whose 
distance  from  the  centre  is  equal  to  the  given  angle,  inter- 
secting CPD,  in  Q.  From  Q  as  a  projected  pole,  describe 
the  circle  EOF,  which  makes  the  required  angle  OEA. 

CoR. — Conversely  BL  is  the  measure  of  the  angle,  AEO. 


PROBLEM  VII. 


To  project  a  great  circle,  which  shall  make  given  angles 
with  two  given  circles. 


1 .   When  one  of  the  given  angles  is  a  right-angle. 

Fig-  3.  Let  the  Primitive  and  the  right  circle  DFB,  be  the  giv- 
en circles.  It  is  required  to  project  a  great  circle  which 
shall  make  right  angles  with  DB,  and  any  given  angle  with 
the  Primitive.  Draw  AC  perpendicular  to  DB.  Set  off 
the  given  angle  from  D  to  E,  project  E  in  F,  and  CFA,  i? 
the  circle  required,  F  and  A  the  required  angles. 

2.    When  neither  of  the  angles  are  right. 

Fig.  3'2.      Let  the  Primitive  and  ABd  be  the  given  circles. 

Find  q  the  pole  of  ABt/,  around  it  describe  a  small  circle 
at  the  distance  of  the  angle  B.  (Prob.  11.)  Set  the  angle 
C  from  d  to  6,  project  b  in  a,  around  the  centre  or  pole  of 
the  Primitive,  describe  a  small  circle  at  the  distance  of  a, 
intersecting  the  other  small  circle  in  s,  which  is  the  pole  af 
the  circle  required,  viz,  CBe  which  describe  by  Prob.  I- 


PROBLEM  VIII. 

^'^To  set  off  any  number  of  degrees  on  a  projected  circle,^^ 
Or,  to  cut  off  from  a  projected  circle  an  arc.  v^hich  corres- 
ponds to  an  arc  of  the  circle  to  be  projected  equal  to  a 
given  arc. 


APPENDIX  TO  SPHERICAL  GEOMETRY.  Igg 

i .    fVhen  the  projected  circle  is  the  Primitive,  or  one  of  its 
parallels. 

Make  DG,  equal  to  the  given  arc,  it  is  the  arc  required,  ^^o-  15. 

EF  on  a  parallel  circle,  is  an  arc  of  the  same  "  number 
of  degrees,''  or  it  is  the  same  portion  of  the  circle  EFX, 
that  DG  is  of  DAB. 


2.   When  the  projected  circle  is  a  right  circle. 

Let  the   projected  circle    be  ACB ;  make  AE  or  DH,  Fig.  lO 
RN  or  RX,  equal  to  the  given  arc;  then  AF  or  CG,  OH  ^  ^^• 
or  OG  will  be  the  required  arcs. 

If  TU  be  made  equal  to  the  given  arc,  then  OB  will  be  Fig.  18. 
the  required  arc,  on  the  great  circle,  and  XY,  a  proportion- 
al arc,  on  the  parallel  small  circle  MN. 


3.   When  the  projected  circle  is  oblique. 

Let  the   projected  circle   be  CSD.     Find  its   pole  P.  Fig.  12- 
(Prob.  L  Cor.)     Draw  PSN,  from  N,  cut  ofFNB  or  NX 
equal  to  the  given   arc  ;  draw    PB    and   PX,  intersecting 
CSD  in  R  and  O  ;  SR  or  SO,  are  the  arcs  required. 

Here  PX  and  PB,  represent  a  small  circle  passing  through 
the  remotest  poles  of  the  Primitive,  and  of  the  oblique  cir- 
cle CSD.     (Sp.  Geom.  HL  Prop.  VHL  Cor.  2.) 

If  CPA,  represent  the  given  oblique  circle,  and  NL  the  Fig.  19. 
given  arc,  then  OQ  will   be  the  required  arc    on  the  great 
circle,  and  XY  the  corresponding  arc  on  a  small  circle  par- 
allel to  it. 

CoR. — Conversely,  if  the  arc  of  a  projected  circle  be 
given,  the  corresponding  arc,  on  the  circle  to  be  projected 
mav  be  found  and  measured. 

thus  AE  and  DH,  RN  and  RX  are  the  measures  of  AF  Fig.  m 
and  CG,  OH  and  OG.  ^  ^^• 

Also   TL^,    measures    OB    and  XY,  and  LN  measures  is  A-  ly 
OQandXY. 


170  APPENDIX  TO  SPHERICAL  GEOMETRY. 

PROBLEM  IX. 

To  project  a  Hemisphere  on  the  plane  of  (he  Equator 

VL  X\V.      Let  the  Primitive  ENWS,  represent  the  Equator. 

Draw  the  perpendicular  diameters  EW,  NS. 

Divide  each  quadrant  into  nine  equal  parts,  then  the 
lines  drawn  from  their  division  to  the  centre,  will  be  the 
projection  of  Meridians. 

Project  parallels  of  Latitude,  corresponding  to  the  divi- 
sions of  the  Primitive,  (Prob.  II.)  also  the  Tropic  ami  Po- 
hr  circles,  at  the  proper  distances  from  the  centre  or  pole 
of  the  Primitive  ;  The  Ecliptic  may  be  projected  by  Proh. 
IV. 

PROBLEM  X. 

^'To  project  a  Hemisphere  on  the  plane  of  the  Meridian. 

Describe  the  Primitive  and  divide  it  as  before.  Let  it 
Represent  the  solstitial  colure. 

WE  is  the  projected  Equator.  Project  parallels  to  it, 
with  the  Tropics  and  Polar  circles,  by  Prob.  II. 

Tangents  to  the  Primitive,  at  the  several  divisions,  will 
intersect  NS  produced  in  the  centres  of  their  parallel  cir- 
cles, as  in  y,  y. 

The  meridians  may  be  projected  by  Prob.  IV.  (iheii 
centres  are  all  in  WE  produced.) 

The  projection  of  the  Ecliptic  is  obvious. 


SPFiERICAL  TRIGONOMETRY, 

PART  I. 


Geometrical  Principles  of  Spherical  Trigonome- 
try ;  or  the  mutual  relations  of  the  Trigonomerriral  lines, 
corresponding  to  the  arcs  and  angles  of  Spherical  Triangles. 


PROPOSITION  I. 

In  right  angled  spherical  triangles,  the  sine  of  either  ol 
the  sides  about  the  right  angle,  is  to  the  rauius  of  the 
sphere,  as  the  tangent  of  the  remaining  side  is  to  the  tan- 
gent of  the  angle  opposite  to  that  side. 

Let  ABC   be   a   triangle  having   the  right   angle  at  A.  PI-  X. 
Then,  sine  of  AB  :  rad:  ".tang.  AC  :  tang.  ABC.  Y\%.  24. 

Lei  D  be  the  centre  of  the    sphere;  join  AB,  AD,  AC, 
and  let   AF  be  drawn  perpendicular  to  BD,  which  there- 
fore will  be  the  sine  of  the  arc  AB  ;  and   from  the  point  F, 
let  there  be  drawn  in  the  plane  BDC  the  straight  line  FE 
at  right  angles  to  BD,  meeting  DC  in  E,  aad  let   AE  be 
joined.     Since  therefore  the  straight  line  DF  is  at  right  an- 
gles to  both  FA  and  FE,  it  will  also  be  at  riiiht  angles  to 
the  plane  AEF,  wherefore  the  plane  ABD,  which  passes^up.  ^2. 4 
through  DF  is  perpendicular  to  the  plane    AEF,  and   the  "    "^ 
plane   AEF,  is  perpendicular   to   ABD  :  But   the    plane 
ACD  orAED.  is  also  perpendicular  to  the  same  ABD  be- 
cause the  spherical  angle  BAC  is  a  right   an^le  :  There- 
fore AE,  the  common  section  of  the  planes  y^ED,  AEF  is    P'  *    ' 
it  at  right  angles  to  the  plane  ABD  and   EAF,  EAD  are 
right  angles. 


i72  SPHERICAL  TRIGONOMETRY. 

Therefore  AE  is  the  tangent  of  the  arc  AC  and  in  the 
rectilineal  triangle  AFE  having  aright  angle  at  A, 

AF  :  rad.::AE  :  tang.  AFE, 
but   AF=sine  ofAB,  and  AE=tang.  of  AC,  and  AFE= 

[ABC. 
.'.     sin  AB  :  rad.:  :tang.  AC  :  ABC. 

Q.  E.  D. 


Cor. — Since  sin    AB  :  rad:  :tang.  AC  :  tang.  ABC, 
and  (PI.  Trig.  93.)  R  :    cot.  ABC: : tang.  ABC  :  rad, 
.-.     Sin.  AB  :  cot.  ABC: : tang.  AC  :  rad. 


PROPOSITION  II. 

In  right  angled  Spherical  Triangles,  the  sine  of  the  Hy- 
potenuse is  to  the  radius,  as  the  sine  of  either  side  is  to  the 
sine  of  the  angle  opposite  to  that  side. 
Fig.  25.      That  is,  sin  BC  :  rad :;  sin  AC  :  sin  ABC. 

Let  D  be  the  centre  of  the  sphere,  and  let  CE  be  drawn 
perpendicular  to  DB,  which  will  therefore  be  the  sine  of 
the  Hypotenuse  BC :  and  from  the  point  E  let  there  be 
drawn  in  the  plane  ABD  the  straight  line  EF,  perpendicu- 
lar to  BD,  and  let  CF  be  joined  :  then  CF  will  be  at  right 
Sup.2.i8.angles  with  the  plane  ABD. 

Wherefore  CFD,  CFE  are  right  angles,  and  CF  is  the 
sine  of  the  arc  AC  :  and  in  the  triangle  CFE,  having  the 
right  angle  CFE, 

CE  :rad::CF  :  sin  CEF, 
ButCEF=ABC,  (Spher.  Geom.  II  Def.  l.cor.  5.) 
.'.  sin.  BC  :  rad: : sin  AC  ;  sin  ABC. 

Q.  £.  D, 


SPHERICAL  TRIGONOMETRY.  173 


PROPOSITION  HI. 


In  right-angled  spherical  triangles,  the  cosine  of  the 
Hypotenuse  is  to  the  radius,  as  the  cotangent  of  either  of 
the  angles  is  to  the  tangent  of  the  remaining  angle. 

That  is,  cos.  BC  :  rad.:  :cot.  ABC  :  tan.  ACB.  Fig.  56. 

Describe  the  circle  DE,  of  which  B  is  the  pole,  and  let 
it  meet  AC  in  F,  and  the  circle  BC  in  E  ;  and  since 
the  circle  BD  passes  through  the  pole  B,  of  the  circle  DF, 
DF  must  pass  through  the  pole  of  BD. 

And  since  AC  is  perpendicular  toBD,  (Sp.  Geom.  Def, 
9,  cor.  5.)  therefore  AC  must  also  pass  through  the  pole  of 
BAD,  wherefore  the  pole  of  the  circle  BAD  is  in  the  point 
F,  where  the  circles  AC,  DE  intersect. 

The  arcs  FA,  FD  are  therefore  quadrants,  and  likewise 
the  arcs  BD,  BE.  Therefore,  in  the  triangle  CEF  right- 
angled  at  the  point  E,  CE  is  the  complement  of  BC,  the 
Hypotenuse  of  the  triangle  ABC  ;  EF  is  the  complement 
of  the  arc  ED,  the  measure  of  the  angle  ABC;  and  FC, 
the  Hypotenuse  of  the  triangle  CEF,  is  the  complement  of 
AC;  and  the  arc  AD,  whicli  is  the  measure  of  the  angle 
CFE,  is  the  complement  of  AB. 

But  (I)Sin.CE  :  R::tan.  EF  :  tan.  ECF, 

That  is,  COS.  BC  :  R : :  cot.  ABC  :  tan.  ACB. 

q.  JE.  D. 

CoR.— Since  cot.  ACB  :  R:  :R  :  tan.  ACB,  i^i-  Trig. 

.-.    cot.  ACB  :  COS.  BC:  :R  :  cot.  ABC.      ^^• 


PROPOSITION  IV, 


In  right-angled  spherical  triangles,  the  cosine  of  an  an- 
gle is  to  the  radius,  as  the  tangent  of  the  side  adjacent  to 
that  angle,  is  to  the  tangent  of  the  Hypotenuse. 

That  is,  cos.  ABC  :  rad. :  itang.  AB  :  tang.  BC.  Fig.  26. 

2.3 


174  SPHERICAL  TRIGONOMETRY. 

For  (I)  sine.  FE  :  rad.:  itang.  CE  :  tan.  CFE, 
But   sine   EF  =  cos.    ABC,   tang.  CE  =  cot.  BC,   and 
tang.  CFE  =  cot.  AB. 

.- .     COS.  ABC  :  rad. : !  cot.  BC  :  cot.  AB. 


Again  cot.  BC  :  rad.:  irad.  :  tang.  BC, 
And  cot.  AB  :  rad.:  irad.  :  tang.  AB, 
.'.     cot.  BC  :  cot.  AB: : tang.  AB.  :  t; 
Therefore,  cos.  ABC  :  rad. : :  tang.  AB  :  1 


:  cot.  AB:  :tang.  AB.  :  tang   BC; 
ABC  :  rad. : :  tang.  AB  :  tang.  BC. 


q.  E.  D. 


CoR.  1.  —From  the  demonstration  it  is  manifest,  that  the 
tangents  of  any  two  arcs  AB,  BC  are  reciprocally  pro- 
portional to  their  cotangents. 

CoR.  2  —Cos.  ABC  :  cot.  BC:  :tan.  AB  :  rad. 


PROPOSITION  V. 

In  right-angled  spherical  triangles,  tne  cosine  of  either 
of  the  sides  is  to  the  radius,  as  the  cosine  of  the  Hypote- 
nuse, is  to  the  cosine  of  the  other  side* 

That  is,  cos.  CA  :  rad.  :  cos.BC  :  cos.  AB. 

For  (II)  Sin.  CF  :  rad. :  :sin.  CE  :  sin.  CFE, 
ButSin.CF=cos.CA,  sin.CE=cos.BC,and  sin.CFE  = 

[cos  AB. 
.'.  cos.  A  :  rad. :  :cos.  BC  :  cos.  AB. 


Q.  E.  D. 


PROPOSITION  VI. 


In  right-angled  spherical  triangles,  the  cosine  of  either 
of  the  sides  is  to  the  radius,  as  the  cosine  of  the  angle  op- 
posite to  that  side,  is  to  the  sine  of  the  other  angle. 

That  is,  cosine  CA  :  rad  : :  cosine  ABC  :  sin.  BCA. 


SPHERICAL  TRIGONOMETRY.  J75. 

For  (11)  sin.  CF  t  rad.::sin.  EF  :  sin.  ECF, 

But     ''     sin.  CF=cos.  CA,  sin.  EF=cos.  ABC,  and  sin. 

ECF=sin  BCA. 
^.  COS.  CA  :  rad.:  .cos. ABC  :  sin.  BCA. 

Q.  E.  D, 


PROPOSITION  VII. 

In  spherical  triangles,  whether  right-angled,  or  oblique- 
angled,  the  sines  of  their  sides  are  proportional  to  the 
sines  of  the  angles  opposite  to  them. 

That  is,  sin.  AC  :  sin.  B:  isin.  AB  :  sin.C  .  Fi?.  13. 

1.  For  (II)  sin.  BC  :  rad.  (=sin.  A.):  isin.  AC  :  sin.  B, 
Also    *'    sin.  BC  :  sin.  A:  :sin.  AB  :  sin.  C, 

.:.     "    sin.  AC  :  sin.  B:  :sin.  AB  :  sin.  C. 

2.  Then,  sin.  BC  :  sin  AC:  Isin  A:  .'sin  B,  Fig.  14. 
Through  C,  draw  a  perpendicular  CD  to  the  opposite    ^  ^^' 

^ide,  then 

(II)  sin.  BC  :  rad.:  :sin.  CD  :  sin.  B, 
"  sin.  AC  :  rad.:  :sin.  CD  :  sin.  A. 
.:.    sin.  BC  :  sin.  AC:  :sin.  A  :  sin.  B. 

In  like  manner,  sin.  BC  :  sin.  AB:  :sin.  A  :  sin.  C. 

Q.  E,  D, 


PROPOSITION  VIII. 


In  oblique  angled  spherical  triangles,  a  perpendicular 
arc  being  drawn  from  any  of  the  angles   on  the  opposite 
side,  the  cosine  of  the  angles  at  the  base,  are  proportion-  Fi^.  u. 
al  to  the  sines  of  the  segments  of  the  vertical  angle. 

Let  ABC  be  a  triangle,  and  the  arc  CD  perpendicular 
to  the  base  BA 
Then,  cos.  B  :  cos.  A:  :sin.  BCD  :  sin.  ACD. 


176  SPHERICAL  TRIGONOMETRY. 

For  (VI)    COS.  CD  :  rad.:  :cos.  B  :  sin.  DCB, 
and      "      COS.  CD  :  rad.!  :cos.  A  :  sin   ACD, 

COS.     B  :  COS.  A:  :sin.  DCB  :  sin  ACD. 

Q,  E.  D. 


PROPOSITION  IX. 

The  same  things  remaining,  the  cosines  of  the  sides  are 
proportional  to  the  cosines  of  the  segments  of  the  base* 

rig.  14.     That  is,  COS.  BC  :  cos.  AC :  :cos.  BD  :  cos.  AD. 

For  (V)    cos.  BC  :  COS.  BD:  :cos.  DC  :  rad. 
And  "      COS.  AC  :  cos.  AD:  :cos.  DC  :  rad. 

COS.  BC  :  cos.  BD::cos.  AC  :  COS.  AD, 
And  COS.  BC  :  cos.  AC:: cos.  BD  :  cos.  AD. 

Q.  E,  D, 


PROPOSITION  X. 

The  same  construction  remaining,  the  sines  of  the  seg- 
ments of  the  base  are  reciprocally  proportional  to  the  tan- 
gents of  the  angles  at  the  base. 

That  is,  sin.  BD  :  sin.  AD: : tan.  A  :  tan.  B. 

For  (I)  sin.  BD  :  rad.: : tan.  DC  :  tan.  B, 
and  "  sin.  AD  :  rad.:  :tan.  BC  :  tan.  A, 
.*.     "     sin.  BD  :  sin.  AD : :  tan.  A  :  tan  B. 

Q.  E.  D. 


PROPOSITION  XI. 

The  same  construction  remaining,  the  cosines  of  the 
segments  of  the  vertical  angle,  are  reciprocally  propor- 
tional to  the  tangents  of  the  sides. 


SPHERICAL  TRIGONOMETRY.  1 77 

That  is,  COS.  BCD  :  cos.  ACD:  :tan.  AC  :  tan.  BC. 

For  (IV)  COS.  BCD  :  R:  :tan.  CD  :  tan.  BC, 
And      "     COS.  ACD  ;  R:  :tan.  CD  :  tan.  AC, 

COS.  BCD  :  COS.  ACD:  :tan.  AC  :  tan.  BC. 

Q.  E.  2>. 


PROPOSITION  XII. 

The  tangent  of  half  the  sum  of  the  segments  of  the  base, 
is  to  the  tangent  of  half  the  sum  of  the  sides,  as  the  tangent 
of  half  the  difference  of  the  sides,  is  to  the  tangent  of  half 
the  difference  of  the  segments  of  the  base. 

Then, tan.  i  BD  +  aD  : tan.  ^BC-fAC : :  tan.  i  BC-  AC 
:  tan.  iBD-AD. 

For  (IX.)  COS.  BC  :  cos.  AC:  :cas.  BD-f  cos.  AD, 
*.  cos  BC+cosAC  :  COS.  BC— cos.  AC::cos.  BD-f-cos. 

AD  :  COS.  BD-cos.  AD. 
But   cos.     BC+cos.    AC  :  COS.    BC~cos.    AC::cot.     | 

BC-fAC  :  tan.  |  BC-AC,* 
Also,    COS.   BD-f  COS.    AD  :  cos.  BD-cos.  AD::cot.    | 

BD-f  AD  :  tan.  i  BD-AD. 


cot.  iBB-fAC  Man.  i  BC  -  AC:  :cot.  1  BD-f  AD  : 


tan.  iBD-AD. 
But,  tan.  i  BC-fAC  X  cot.  i  BC-fAC  :  tan.  1  BC-fAC 
Xtan.  1  BC-AC::tan.  i  BD-f  ADXcot.  i  BD-f  AD 
:  tan.  1  BD-f  ADxtan.  i  BD-AD, 


But,  tan.  I  BC-fAC Xcot.    1  BC-f AC=rad.=tan.    I 
BD-f  AD  cot.  1  BD+AD, 


tan.  i  BC-f  AC  Xtan.  i  BC-AC=tan.  i  BD-f  AD 
Xtan.  i  BD-AD, 

*  Plaue  Trigf. 


178  SPHERICAL  TRIGONOMETRV. 


Or  tan.  i  BD  +  AD  :  tan.  i  BC4-AC:  :tan.  IBC-AC  : 
tan.  iBD-AD. 

Q.  E.  D, 

ScHOL. — The  preceding  proposition  will  be  easily  re- 
membered, from  its  resemblance  to  the  corresponding 
proposition  in  Plane  Trigonometry. 


PROPOSITION  XIII.    DAT. 


tfthe  ratios  of  the  radius  of  any  circle,  to  the  trigonomet- 
rical lines  corresponding  to  any  given  arc,  were  known,  and 
conversely;  then  if  any  two  of  the  parts  of  a  ri^ht-angled 
spherical  triangle  were  given,  the  remaining  parts  would 
also  be  given,  i,  e.  they  might  be  found. 

I.  If  the  hypotenuse  and  either  of  the  adjoining  angles 
were  given,  the  sides  and  the  remaining  angle,  would  be 
PI.  XII.  given  also. 

Fig.  21.      That  is,   if  AC,  and   A  are  given,   AB,   BC,   and  C 
may  be  found. 

1.  Because  the  radius,  the  sine  of  AC,  and  the  sine  of 
A  are  given,  the  sine  of  BC  and  consequently  the  arc  BC, 
may  be  found. 

For  (II)  R  :  sine  AC::  sin.  A  :  sin.  BC ;  and  a  fourth 
6.  12.    proportional  to  any  three  straight  lines  may  be  found. 

Therefore  the  sine  of  BC  may  be  found. 

But  as  the  radius,  and  the  sine  of  BC  are  given,  their 
ratio  is  given,  and  therefore  by  supposition  the  arc  BC  is 
given  also. 

It  is  to  be  remarked  that  any  sine  may  be  the  sine  of 
two  arcs  which  are  supplemental  to  each  other,  but  as  A 
is  given,  it  is  known  of  what  affection  it  is,  i.  e.  whether  it 
is  greater  or  less  than  a  right-angle,  and  BC  is  of  the  same 
affection,  (Sp.  Geom.  XIV.)  therefore  BC  is  given  without 
ambiguity. 


SPHERICAL  TRIGONOMETRY.  I79 

2.  In  a  similar  manner  AB,  may  be  found. 
For  (IV)  R  :  COS.  A:  :tang.  AC  :  tang.  AB. 

If  AC  be  less  than  a  quadrant,  AB  is  of  the  same  affec- 
tion as  A  which  is  given.  If  AC  be  greater  than  a  quad- 
rant, AB,  and  A,  are  of  different  affections,  in  either  case 
AB  is  given  without  ambiguity.  (Sp.  Geom.  XIV.  XV.) 

3.  Also,  C  may  be  found. 

For  (III)  R  :  cos.  AC:  :tang.  A  :  cot.  C. 

If  AC  is  less  than  a  quadrant,  C  is  of  the  same  affectioa 
as  A;  otherwise  they  are  of  different  affections.  In  either 
case  C  is  given  unambiguously. 

In  the  same  manner,  if  AC  and  C  are  given,  BC,  AB, 
and  B  may  be  found. 

II.  If  the  hypotenuse  and  either  of  the  sides  are  given 
the  angles  and  the  remaining  sides  are  given. 

That  is,  if  AC,  BC,  are  given,  AB,  A  and  C  may  be 
found. 


1 .  The  side  AB  may  he  found. 

For  (V.)  cos.  BC  :  cos.  AC:  :R  :  cos.  AB. 

If  AC  is  less  than  a  quadrant,  AB  is  of  the  same  affec- 
tion as  BC,  if  not,  they  are  of  different  affections. 

2.  The  angle  A  may  be  found. 

For  (11)  sine  AC  :  sine  BC : :  R  :  sin.  A. 
The  angle  A  is  of  the  same  affection  as  BC. 

3.  The  angle  C  may  be  found. 

For  (IV)  tangent  AC  :  tan.  BC:  :R  :  cos.  C. 

If  AC  is  less  than  a  quadrant,  BC  and  C,  are  of  the 
same  affection,  otherwise  they  are  of  different  affections. 

In  the  same  manner,  if  AC  and  AB  are  given,  BC,  A  and 
^  may  be  found. 


180  SPHERICAL  TRIGONOMETRY. 

III.  Jf  one  side  and  an  adjacent  angle  are  given,  the  hy- 
potenuse and  the  remaining  side  and  angle,  are  given. 

That  i»,  if  BC  and  C  are  given,  AC,  AB  and  A  may  be 
found. 

1.  The  hypotenuse  JlC  may  be  found* 
For  (IV)  COS.  C  :  R:  :tan  BC  :  tan.  AC. 

If  BC  and  C,  (both  of  which  are  given,)  be  of  the  same 
affection,  AC  is  less  than  a  quadrant,  otherwise  it  is 
greater. 

2.  The  side  AB  may  he  found. 
For(I.)R  :  sin.  BC::tan.  C  :  tan  AB. 
AB  is  of  the  same  affection  as  C. 

3.  The  ansrle  A  may  he  found. 

For  (VI)  k  :  cos.  'BC:  :sin.  C  :  cos.  A, 
The  angle  A  is  of  the  same  affection  as  BC. 

In  like  manner,  if  AB  and  A  are  given,  AC,  BC  and  C 
may  be  found. 

IV.  If  one  side,  and  the  angle  opposite  are  given,  the 
hypotenuse  and  the  remaining  side  and  angle  are  given. 

That  is,  if  BC  and  A  are  given,  AC,  AB  and  B  may  be 
found. 

1.  The  hypotenuse  AC  may  be  found. 
For  (II)  sin.  A  :  sin.  BC : :  R  :  sin.  AC. 

The  hypotenuse  may  be  either  the  arc  AC,  or  its  sup- 
plement Cd^  for  both  have  the  same  sine. 

The  two  triangles  ABC,  and  BCc/  have  each  the  parts 
given  in  this  case.  The  hypotenuse  therefore  is  given 
ambiguously.  It  is  one  of  two  given  arcs.  If  the  affec- 
tion of  either  of  the  remaining  parts  of  the  triangle,  should 
be  given,  the  ambiguity  would  be  removed. 

2.  The  side  AB  may  he  Jound, 
For(l)  tan.  A  :  tan.  BC::R  :  sin  AB. 

The  arc  is  ambiguous,  being  either  AB  or  its  supple- 
ment B(f.     The  sine  of  both  being  the  same. 


SPHERICAL  TRIGONOMETRY. 

3.   The  angle  C  may  he  found. 

For,  (VI)  COS.  BC  :  cos.  A:  :R  :  sin.  C,  which  is  ambi- 
guous,  being  either  BCA,  or  BCJ,  which  have  the  same 
sine. 

In  like  manner  if  AB  and  C  are  given,  AC,  BC,  and  A 
may  be  found. 

V.  If  the  two  sides  are  given,  the  hypotenuse  and  an- 
gles are  given. 

That  is,  if  AB  and  BC  are  given,  AC,  A  and  C  may 
be  found. 

1 .    The  hypotenuse  AC  may  he  found. 

For,  (V)  R  :  cos.  AB::cos.  BC  :  cos.  AC,  which  is 
less  than  a  quadrant,  when  AB  and  BC,  are  of  the  same 
affection,  but  greater  than  a  quadrant  when  they  are  of 
different  affections.    (Sp.  Geom.  XV.) 

2.  The  angle  A  may  he  found. 

For,  (I)  sin.  AB  :  R:  :tan.  BC  :  tan.  A  ;  which  is  of  the 
same  affection  as  BC. 

3.  In  the  same  manner  B  may  be  found. 

For,  (I)  sin.BC  :  R::tang.  AB  :  tan.  C  ;  which  is  of  the 
same  affection  as  AB. 

VI.  If  the  two  angles  are  given,  the  Hypotenuse  and  the 
opposite  sides  are  also  given. 

That  is,  if  A  and  C  are  given,  AC,  AB,  and  BC  may  be, 
found. 

1.  The  hypotenuse  AC  may  be  found. 

For,  (HI)  tan.  A  :  cot.  C:  :R  :  cos.  AC. 
When  B  and  C  are  of  the  same  affection,  AC  is  less  than 
a  quadrant;  but  when  they  are  of  diffi^reiU  affections,  AC 
is  greater  than  a  quadrant.     (Sp.  Geom.  XIV.  XV.) 

2.  The  side  AB  may  be  found. 

For  (VI)  sin.  A  :  cos.  C:  :R  :  cos.AB  ;  which  is  of  the 
same  affection  as  C. 

24 


181 


182  SPHERICAL  TRIGONOMETRY. 

3.  In  the  same  manner  BC  may  he  found. 
For,  (VI)  sin.  C  :  cos.  A:  :R  :  COS.  BC,  which  is  of  th« 
same  affection  as  A. 


PROPOSITION  XIV.  DAT. 

The  same  being  supposed  as  before,  if  any  three  of  the 
parts  of  an  obhque  angled  triangle   be  given,    the  other 
three  are  also  given, 
^'o  32.      I,  jf  ^^Q  sides  artd  the  included  angle  afe  given,  the  re- 
maining side  and  angles  are  also  given. 

That  is,  if  AC  and  CB  and  C,  are  given,  AB  and  A  and 
B  ;  may  be  found. 

1.  The  angle  B,  may  be  found. 

For  let  fall  the  perpendicular  AP,  from  the  other  un- 
known angle  A,  on  BC, 

Then,  (IV)  R  :  cos.  C:  :tan.  AC  :  tan.  CP  ;  which  is 
therefore  given. 

And  (X)  sin.  BP  :  sin.  CP:  :tan.  C  :  tan  B  ;  which  is  of 
the  same  affection  as  A  when  the  perpendicular  falls  within 
the  triangle,  but  if  it  falls  without,  B  and  A  are  of  differ- 
ent affections. 

In  like  manner,  the  angle  A  may  be  found  ;  or  each  of 
its  parts  CAP,  and  PAB,  may  be  found  ;  as  in  case  HI, 
Prop.  XIII  Dat. 

2.  The  side  AB  may  be  found. 

For  (IV)  R  :  COS.  C:  :tan.  AC  :  tan.  CP, 
And  (IX)  COS.  CP  :  COS.  BP:  :cos.  AC  :  COS.  AB  ;  which 
is  of  the  same  affection  as  AC,  when  AP  falls  within  the 
triangle,  but  different  when  it  falls  without.     (Sp.  Geom. 
XVI.) 
Or  AB,  may  be  found,  after  B  is  found,  by  Prop.  VII. 

II.  If  two  angles  and  the  included  line  are  given,  the  re-" 
maining  sides  and  angle  are  given. 

That  is,  if  A,  C,  and  AC,  are  given^  AB;  CB  and  B  may 
be  found. 


SPHERICAL  TRIGONOMETRY.  183 

I.  If  AB  may  he  found. 

For,  letting  fall  the  perpendicular  as  before, 

Then  (III)  R  :  COS.  AC:  :tan.  C  :  cot.  CAP, 

Therefore  BAP  may  be  found  (=CAB-CAP,) 

which  is  of  the  same  affect  as  C,  when  AC  is  less  than 
a  quadrant. 

and  (XI)  COS.  BAP  :  cos.  CAP  :  :tan.  AC  :  tan.  AB. 

AB,  is  of  the  same  affection  as  AC,  when  the  perpen- 
dicular falls  within  the  triangle,  i,  e,  when  CAP<CKB.  ! 

In  like  manner,  BC  may  be  found. 
Or,  after  AB  is  found,  by  Prop.  VII. 

2.   TTie  third  angle  B  may  be  found. 

For,  (III)  R  :  cos.  AC: : tan.  C  :  cot.  CAP. 

and  (VllI)  sin.  CAP  :  sin.  BAP::cos.  C  :  cos.  B; 
which  is  of  the  same  affection  as  C,  when  AP  falls  within  the 
triangle,  but  they  are  of  different  affections,  when  AP  falls 
without. 

Or,  when  AB  is  found,  B  may  be  found,  by  Prop.  VII. 

III.  If  two  sides,  and  an  angle  opposite  to  one  of  them 
be  given,  the  remaining  angles  and  side  are  given. 

That  is,  if  AC  and  AB  and  C,  be  given,  B,  A,  and   BC  • 

may  be  found. 

1.  TTie  angle  B  opposite  AC  may  be  found. 
For  (VII)  sin.  AB  :  sin.  AC:  :sin.  C  :  sin.  B. 
When  AB-f-AC[>C6,C>AB6, 

But,,  AB-f-AC<Ce,  C<  ABe,  (Sp.  Geom.  XII.  h  Cor.) 
This  proposition  may  sometimes  determine  whether  B  is 
acute  or  obtuse  ;  when  it  does  not,  B  is  ambiguous. 

2.  The  angle  A  is  given.  . 
For,  the  perpendicular  being  drawn  from  A, 

Then  (III)  R  :  COS.  AC : :  tan.  C  :  cot.  CAP. 

And  (XI)  tan.  AB  :  tan.  AC: : cos.  CAP  :  cos.  BAP. 

Therefore  CAP  and  BAP  being  given,  CAB  is  given^ 
hut  it  may  be  either  acute  or  obtuse,  as  the  perpendicular 
falls  within  or  without  the  triangle. 

Tt  is  therefore  ambiguous. 


)g4  SPHERICAL  TRIGONOMETRY 

3.    The  third  side  BC  is  given. 
For.  (IV)  R  :  COS.  C:  itan.  AC  :  tan.  CP. 
And  (IX)  COS.  AC  :  cos.  AB:  :cos.  CP  :  COS.  BP. 
CB  may  therefore  be  found,  but  it  is  ambiguous,  as  AP, 
may  fall  within  ov  without  the  triangle. 

Or,  after  A  is  given,  Be  may  be  found,  by  Prop.  VII. 

IV.  If  two  angles  and  a  side  opposite  lo  one  of  them  be 
given,  the  remaining  sides  and  angle  are  given. 

That  is,  if  C,  B  and  AC,  be  given.  A,  BC,  and  AB  may 
be  found. 

1.  The  side  AB  opposite  C  may  be  found* 

For,  (VII)  sin.  B  :  sin.  AC: : sin.  C  :  sin.  AB. 
When  C4-B>2  right-angles  AC -1-AB>C^ 
and„C-hB<2     „        „     AC+AB<Ce.  (S.  G.  XII.) 
This  proposition  will  sometimes  determine  whether  AB 
is  greater  or  less  than  a  quadrant,  otherwise  it  is  ambigu- 
ous. 

2.  The  side  BC,  adjacent  to  C  and  B  may  be  found. 
For,  (I)  R  :  cos.  C:  :tan.  AC  :  tan.  CP', 

and,  (X)  tan.  B  :  tan.  C:  :sin.  CP  :  sin.  BP. 
BP  is  ambiguous  (see  Prop.  XIIl,  Dat.  case  IV.) 

Therefore  BC,  is  ambiguous  both  when  the  perpendic- 
ular falls  within  the  triangle  and  when  it  falls  without,  so 
that  BC  might  have  four  values,  except  that  some  of  them 
are  excluded  by  the  impossibility  of  its  being  greater  than 
a  semi-circle. 

S.   The  remaining  angle  A  may  be  found. 
For  (III)  R  :  cos.  AC: : tan.  C  :  cot.  CAP. 
And  (VIII)  cos.  C  :  COS.  B:  :sin.  CAP  :  sin.  BAP. 
A  is  ambiguous,  in  the  same  sense  as  BC. 

V.  If  the  three  sides  are  ^iven,  the  angles  may  be  found. 
That  is,  if  AC,  AB  and  BC,  be  given,  A,  B,  and  C  may 

be  found. 


SPHERICAL    TRIGONOMETRY.  185 

I.   The  unsle  C  may  be  found. 

For,  (XII)  tan.  iBC  :  tan,  ^AC+AB::tan.  fAC-AB, 
:  tan.  iE  =  the  difference  between  CP  and  BP. 
When  AP  falls  within  the  triangle. 
Then  (I)  tan.  AC  :  tan.  CP:  :R  :  cos.  C. 

In  the  same  manner,  the  other  angles  may  be  found. 
But  after  C  is  found,  the  sides  may  be  found  by  Prop.  VII. 

If  AC-f  AB>Ce,C-}-B>2  right-angles,  and  the  greater 
side,  is  opposite  to  the  greater  angle. 

VI.  If  the  three  angles  of  a  spherical  triangle  are  given, 
the  three  sides  are  given. 

The  supplements  of  the  angles,  are  the  sides  of  the  sup- 
plemental triangle,  (Sp.  Geom.  X.)  which  are  there- 
fore given,  therefore  its  angles  are  given  by  the  last  ;  but 
the  supplements  of  those  angles  are  the  sides  of  the  sup- 
posed triangle  which  therefore  are  given. 

ScHOL.  If  the  sides  of  a  spherical  triangle  are  reduced 
indefinitely,  they  approach  continually  to  equality  with 
their  sines  and  tangents,  and  to  coincidence  with  them.  i.  e. 
the  limit  of  the  ratio  of  the  sides  to  their  sines,  or  tan- 
gents, is  a  ratio  of  equality. 

If  they  be  considered  as  actually  coinciding  in  their 
evanescent  state,  the  triangle,  becomes  a  plane  triangle, 
and  as  the  properties  of  the  spherical  triangle  belong  to 
it,  whatever  be  the  length  of  its  sides,  the  propositions 
which  have  been  demonstrated  concerning  spherical  trian- 
gles, may  be  transferred  to  plane  triangles,  by  substituting 
the  sides  of  such  plane  triangles  in  the  places  of  the  sines 
or  tangents,  of  the  sides  of  the  spherical  triangle.  It  will 
be  seen,  that  these  results  conform  to  the  Propositions  in 
Plane  Trigonometry. 


NAPIER'S  RULES  OF  THE  CIRCULAR  PARTS. 


The  rule  of  the  Circular  PartSy  invented  by  Napier,  is 
of  use  in  spherical  trigonometry,  by  reducing  all  the  the- 
orems employed  in  the  solution  of  right-angled  triangles  to 
two.  These  two  are  not  new  propositions,  but  are  mere- 
ly enunciations,  which,  by  help  of  a  particular  arrange- 
ment and  classification  of  the  parts  of  a  triangle,  include 
the  first  six  propositions,  with  their  corollaries,  which  have 
been  demonstrated  above.  They  are  perhaps  the  happi- 
est examples  of  artificial  memory  that  is  known. 

Definition  1. — If  in  a  spherical  triangle,  we  set  aside 
the  right  angle,  and  consider  only  the  five  remaining  parts 
of  the  triangle,  viz,  the  three  sides  and  the  two  oblique 
angles,  then  the  two  sides  which  contain  the  right-angle, 
and  the  complements  of  the  other  three,  namely,  of  the 
two  angles  and  the  Hypotenuse,  are  called  the  Circular 
Parts . 
Fig.  13.  Thus  in  the  triangle  ABC  right-angled  at  A,  the  circular 
parts  are  AC,  /iB  with  the  complements  of  B,  BC,  and  C. 
These  parts  are  called  circular  ;  because,  when  they  are 
named  in  the  natural  order  of  their  succession,  they  go 
round  the  triangle. 

Def.  2. — When,  of  the  five  circular  parts,  any  one  is 
taken,  for  the  middle  part,  then  of  the  remaining  four, 
the  two  which  are  immediately  adjacent  to  it,  on  the  right 
and  left,  are  called  the  adjacent  parts  ^  and  the  other  two, 
each  of  which  is  separated  from  the  middle  by  an  adjacent 
part,  are  called  opposite  parts, 

*  This  account  of  Napier's  Circular  Parts,  is  taken  principally  from  Play- 
fair's  Appendix  to  Spherical  Trigonometry. 


NAPIER'S  RULES  OP  THE  CIRCULAR  PARTS. 

Thus,  if  AC,  be  reckoned  the  middle  part,  then  AB  and 
the  complement  of  C,  which  are  contiguous  to  it  on  differ- 
ent sides,  are  called  adjacent  parts  ;  and  the  complements 
of  B  and  BC  are  the  opposite  parts.  In  like  manner  if  AB 
be  taken  for  the  middle  part,  AC  and  the  complement  of 
B  are  the  adjacent  parts,  and  the  complements  of  BC  and 
C  the  opposite.  If  the  complement  of  BC  be  the  middle 
part,  the  complements  of  B  and  C  are  adjacent,  AC  and 
AB  opposite  parts. 


PROPOSITION. 

In  a  right-angled  spherical  triangle,  the  rectangle  of  the 
radius  and  the  sine  o(  the  middle  part,  is  equal  to  the  rec- 
tangle of  the  tangents  of  the  adjacent  parts  ;  or  to  the  rec- 
tangle of  the  cosines  of  the  opposite  parts. 

The  truth  of  the  two  theorems  included  in  this  enunci- 
ation may  be  easily  proved,  by  taking  each  of  the  live  cir- 
cular parts  in  succession  for  the  middle  part,  when  the  gen- 
eral proposition  will  be  found  to  coincide  with  some  one 
of  the  analogies  already  given  for  the  resolution  of  the 
cases  of  right-angled  spherical  triangles. 

If  a  spherical  triangle  have  one  side  a  quadrant,  the  sup- 
plemental triangle  will  be  right-angled  ;  and  as  the  sines, 
tangents,  &£c  of  an  arc  are  the  same  as  those  of  its  supple- 
ment, Quadrantal,  or  Rectilateral  spherical  triangles,  may 
evidently  be  solved  like  Rectangular  Triangles.  The 
same  proposition  of  Napier,  is  applicable  to  both,  if  the 
following  are  taken  as  the  Circular  Parts  in  the  former. 

The  Quadrant  is  in  the  place  of  the  right-angle,  and  is 
not  supposed  to  separate  the  circular  parts,  which  are, 
the  two  angles  adjacent  to  the  Quadrant,  the  complements 
of  the  other  two  sides,  and  of  the  remaining  angle. 


SPHERICAL  TRIGONOMETRY, 

PART  11. 


Calculation  of  the  sides  and    Angles    of  Spheri- 
cal Triangles. 

It  has  been  saidbyPlayfair,  that,  "Trigonometry  is  the 
apphcation  of  numbers  to  express  the  relations  of  the 
sides  and  angles  of  triangles  to  each  other.''  This  is  true 
of  Trigonometry,  as  applied  to  the  actual  calculation  of 
the  sides  and  angles  of  triangles.  The  nature  and  use  of 
this  application  of  numbers  will  be  evident,  by  remarking 
that  the  Data,  Prop.  XIII,  and  XIV.  are  founded  on 
the  supposition  that  the  ratios  of  radius  to  the  Trigono- 
metrical lines  belonging  to  any  given  arc,  are  given.  These 
ratios  however  are  not  given,  and  cannot,  except  in  a  very 
few  cases,  be  expressed  by  numbers.* 

If  however  the  circumference  of  the  circle  be  divided 
into  any  number  of  equal  parts,  as  360,  the  ratios  of  radius 
to  the  trigonometrical  lines,  belonging  to  these  arcs,  may 
be  calculated  to  any  required  degree  of  exactness.  {Days 
Trigonometry,  "  Computation  of  the  canon''\)  This  is 
done,  and  numbers  expressing  the  relations  of  their  parts 
to  radius,  considered  as  unity,  are  arranged  in  tables. 

By  means  of  these,  the  sides  and  angles  of  spherical,  as 
of  plane  triangles  can  be  easily  computed,  from  the  pre- 
ceding propositions. 

*^Ca5noli,Chap.  VI, 


I.  RIGHT  ANGLED  SPHERICAL  TRIANGLES. 


CASE  I. 


Given    the    Hypotenuse    (=55°    8',)    and    one   side  Fig.  21. 
(=32°  12',)  it  is  required  to  find  the  other  parts. 

Project  the  Triangle,*  a(  the  circumference  of  the 
Primitive,  making  {Ap  =  )  A'  =55°  8'  and  {fy=fz  =  ) 
BC  =  32°  12'.  (Prob.  VIII.),  on  circles  described  through 
ACrf,  and  B    y. 

Attht  centre^  make  (m  n=AP=)  AC=Hyp.  and  (xy=) 
BC  =  side,  on  the  obhque  circle  described  through  m  and  Fig.  22. 
C. 

In  the  plane,  from  the  point  A,  make  (y2=ym=)  AC=  a      33, 
Hyp.  on  the  oblique  great   circle    described   through  y  A 
and  C,  and  BC,  (on  the  oblique  circle  described  through 
3cCrf)=given  side. 


1,   To  find  the  other  side,  AB. 

By  Circular  Pa  ts;  R  .  cos.  AC=BC  .  cos.  AB. 

Or,  Prop.  XIII.2  . 1  . ;  cos  EG :  R:  :cos.  AC  :  cos.  AB. 

Cos.  BC  =  32°  12'  av  .  co.  0.0725305. 

R  10 

Cos  AC=55°  G'  9.7571444. 


Cos.  AB=47°  30'  4'  9.8296749. 


*  The  Projections  are  not  given  in  full.  To  a  person  familiar  with  the 
Problems  in  Stereographic  Projections,  a  mere  reference  to  the  steps  of  the 
process,  it  is  presumed,  will  be  sufficient. 

25 


190  SPHERICAL  TRIGONOMETRY. 

'Tlie  side  AB  is  less  than  a  quadrant,  like  BC,  because 
the  Hypotenuse  is  less  than  a  quadrant. 

2.  To  find  the  angle  C. 

Cir.  Paris;  R  .  cos  .  C=tan.  BC  .  cot.  AC, 
Or  prop.(Xni.  2.  3.)  tan.  AC  :  tan.BC::  R:cos.C.=^ 
68°  58'  30".    C  <  90°,  like  BC. 

3.  To  find  the  angle  A, 

Cir.  Parts;  R  .  sin.  BC=sin.  AC  .  sin.  A. 
Or  Prop.  (XIII.  2.  2.)   sin.   AC  :  R:  :sin  BC  :  sin.  A, 
40^30  5".     A  likeBC<  90°. 


CASE  II. 

Given  the  Hypotenuse  (=55°  8')  and  one  angle  (=40** 
30'  5   'to  find  the  remaining  angle,  and  the  sides. 

Project  the  Triangle,  making  A=40°  30'  5"   and  AC= 
55°  8'. 
Fi«',  21.      The  projections  at  the  circumference  and  centre  are  evi- 
24.  dent.     From  a  point,  as  A,  in  the  plane,  project  the  circle, 
-^-  making  A =40°   30'   5'',  (Prob.    IV.)     From   A    cut   off 
AC=Hypotenuse  (VIII)  project  the  circle  jt?Co,  through 
C ;  BC,  AB  are  the  required  sides; 

1.  Cir.  Parts;  R  .  cos.  55°  8'=cot.  40°  30*  5"  .  cot.  Q. 
C<90°,  like  A,  because  AC<90°. 

2.  Sin.  A  .  sin.  AC=^R  .  cos.  BC. 
BC<90°.  hke  A  .  because  AC<90°. 

3.  R  .  cos.  A=cot.  AC  .  tan.  AB. 
AB<90  .  like  C;  because  AC  <  90. 


SPHERICAL  TRIGONOMETRY.  191 


CASE  III. 


Fis:.  ^. 


Given  one  side  (=47°  30'  4")  and  its  opposite  angle 
(=63°  58  30")  to  find  the  other  parts. 

Project  the  triangle,  at  the  circumference. 

Make  AB=47°  30'  4"=Draw  the  perpendicular  diam- 
eters Br,  yz.  Make  ym=^n =63°  58'  30'  and  draw  the 
small  parallel  circle,  (Prob.  II.)  from  /,  as  a  pole,  pro- 
ject the  oblique  circle  AC^R.  ACB,  or  RCx  is  the  tri- 
angle required.     Each  of  them  contain  the  given  parts. 

Jit  the  centre.  Cut  ofF  AB  =  given  side  (Prob.  VHI.)  Fig.  27. 
Project  the  circle  dBe  (Prob.  III.);  around  x,  its  pole, 
describe  the  small  circle,  at  the  distance  of  the  given  angle. 
(Prob.  II.)  From  h,  where  it  intersects  the  primitive, 
as  a  pole,  project  the  right  circle  fad;  then  ACB,  or  its 
supplement  is  the  triangle. 

From  a  point  in  the  plane,  as  .^,  cut  off  AB,  on  a  right  ^^g-  "^S. 
circle,  equal  to  the  given  side.  (Prob.  V^III..)  Project 
the  oblique  circle  arBz,  (Prob.  III.)  find  its  pole  m,  and 
around  it,  describe  the  small  circle,  at  the  distance  of  the 
given  angle;  r,  its  intersection  with  the  circle  xvZj  whose 
pole  is  A,  is  the  pole  of  a  circle  passing  through  A,  making 
the  required  angle  atC. 

1.  R  .  COS.  C=cos.    AB  (=  Rx)  .  sin.  A  or  CRa?  its  Fig.26. 
supplement.     For  the  angle  is  ambiguous. 

2.  Tan.  .  AB  (or  Rx)  .  cot.  C=R    sin.  BC  (or  Ca:.) 
The  side  is  ambiguous.     BC  or  its  sup'  Ca. 

3.  R .  sin.  AB=sin.  C  .  sin.  AC,  or  CR  .  which  is  am- 
biguous. 


^g^  SPHERICAL  TRIGONOMETRY. 


CASE  IV. 

Given  one  side  (=47°  30' 4"),  and  the  adjacent  angle 
(40°  30'  5')  to  find  the  other  parts. 

Fig  21.      Project  the  Triangle,  mMngAB^^I"^  30' 4",  andA= 
2^;  40°  30'  5".     (Prob.IV.) 

1.  R  .  sin.  AB=cot.  A  .  tang.  BC. 
EC  is  of  the  same  affection  as  A. 

2.  Cos.   AB  .  sin.  A=R  .  Cos.  C. 
C  is  of  the  same  affection  as  CB. 

3.  R  .  cos.  A = tang.  AB  .  cot.  AC. 

The  Hypotenu^^e   is  less  then  a  quadrant,  because  AB 
and  A  are  of  the  same  affection. 


CASE  V. 

Given  two   sides  (=47°  30'  4"  and  32°  12')  to  find  the 
other  parts. 


Pig.  21.      Project  the  Triangle,  making  AB  and  BC  equal  to  the 

22. 
23. 


given  sides. 


1.  Cos.  AB  .  cos.  BC=R .  cos.  AC. 
AC<  90,  because  AB  and  BC  are  alike. 

2.  R  .  sin.  AB=tang.  BC .  cot.  A. 
A<90,  likeBC. 

3.  R  .  sin.  BC=tan.  AB  .  cot.  C. 
C<90,likeAB. 


SPHERICAL  TRIGONOMETRY*  193 


CASE  VI. 

Given  the  two  angles  (=40°  30'  5"  and  63°  58'  30")  to 
iind  the  other  parts. 

Project  the  Triangle,  Fig.  39. 

At  the  circumference,  make  A =40  30'  5". 

Find  y  the  pole  of  ACx,  and  around  it  describe  the  small 
circle,  at  the  distance  of  63°  58'  30",  its  intersection  with 
the  primitive  in  S,  is  the  pole  of  the  right  circle  BC,  which 
makes  the  given  angle  at  C,  with  ACx. 

At  the  centre,  make  A=one  of  the  given  angles  about  x  the    *'  30. 
pole  of  CArf,  describe  the  small  circle  yz  at  the  distance  of 
the  other  given  angle  from  x,  their  intersection  is  the  pole 
of  5BC/  making  the  given  angle  atC. 

From  K,  a  point  in  the  plane  of  the  Primitive  ;  make    "  31, 
A=    given   angle,  about  m  the  pole  of  ACx,    describe  a 
small  circle,  at  the  distance  of  the  other  angle,  intersecting 
AB^?  in  p  which  is  the  pole  of  dBCf  making  C=the  other 
^iven  angle. 

1.  Cot.  A  .  cot.  C=R.  cos.  AC. 
AC<90,  because  A  and  C  are  alike. 

2.  R  .  cos.  C=sin.  A  .  cos.  AB. 
AB<90,  like  opposite  angle  C. 

2.  R .  COS.  A=sin.  C  .  cos.  BC. 
BC<90  1ikeA. 


OBLIQUE-ANGLED  TRIANGLES. 


CASE  I. 

Given  two  sides  (=58°,  and  79°  17'  14",)  and  the  angle 
opposite  one  of  them  (=62°'  34  6",)  it  is  required  to  find 
Fig.  32.  the  other  parts. 

Project  the  Triangle,  making  AC  =  58°  and  describe  the 
circle  CBe,  making  C=62°  34'  G",  and  the  small  circle 
kg,  at  the  distance  of  79°  17'  14"  from  A,  ABrf  will  com- 
plete the  triangle. 

Through,  q^  the  pole  of  CBc  project  the  circle  Apqd. 
perpendicular  to  Be. 

1.  Sin.  AB  :  sin.  C:  Isin.  AC  :  sin.  B.  i.  c.  sin.  ABC. 
or  ABe. 

This  case  is  ambiguous, 

2.  R  .  cos.  C=cot.  AC  .  tan.  PC. 
PC<90,  because  C,and  AC<  90. 

3.  R  .  COS.  B=cot.  AB  .  tan.  BP. 
PB<90,  as  before,  and  PC+PB=CB. 

4.  Sin.  AC  :  sin.  B:  isin.  BC  :  sin.  A, 
Which  is  ambiguous. 

The  projection  gives  B  acute,  and  A  obtuse,  but  with 
the  same  things  given,  another  triangle  might  be  project- 
ed, containing  angles  equal  to  the  supplements  of  these. 


SPHERICAL  TRIGONOMETRY.  1^5 

That  two  spherical  triangles  may  have  two  sides,  and 
the  angle  opposite  to  one  of  them  in  each  equal,  while  the 
remaining  side  and  angles  are  unequal  may  be  easily 
shown,  as  in  fig.  6,  where  the  triangles  APB,  APB',  have 
AP,  AB.  or  AB',  and  the  angle  P,  in  each  equal.  The 
projection  of  these  two  triangles  is  given  PI.  XVI,  fig.  2, 
where  the  ambiguity  of  A,  B  and  BC  is  exhibited. 
(BC  =  PC^PB,)  and  A,  B,  are  either  acuta  or  obtuse. 


CASE  2. 

Given  two  sides  (  =  58°.  and  110°.)  and  the  included  an- 
gle, (62°.  34'  6")  to  find  the  other  parts. 

Project  the  Triangle  ;    making  AC  =  58°  C=62°.  34'  6"  Fig,-33. 
and  CB  =  nO°,  by  drawing  the  small  circle  pBr^,   at   the 
distance  of  180°.  — 1 10°.  =  70°.  from  e  as  its  pole. 

Describe  A  Br?,  through  B,  and  APrf  through  the  pole 
of  Ce. 

1.  R.  cos.C=cot.  AC  .  tan.  PC  ;  <90°. 
ThenBC-PC=PB. 

2.  (X;  Sin.  PB  :  sin.  PC:  :tan.  C  :  tan.  B. 

B<  90°.  like  C,  because  AP  is  within  the  triangle. 

.3.  Sin.  B  :  sin.  AC::  sin.  C  :  sin.  AB<90^ 
Because  B  and  BP  are  both  acute. 

4.  Sin.  AC  :  sin.  B::sin.  BC  :  sin.  A  .>90°. 
ForBC-fBA>180°5.:.(S.  G.  XII)B  Ae<C. 


CASE  3. 

Given  two  angles  (=50°,  and  62°    34'  6")  and  the  side 
opposite  to  one  of  them  (79°  1 7'  1 4'')  to  find  the  otherparts. 

Project  the  Triangle  ;  making  BA  =  79°    1 7'    1 4",  the  Fig.  3s. 
angle  B=oO°, ;  also  describe  the  great  circle  ACrf,  making 
with  BC  an  angle  at  C=62°  34'  6" 

Draw  APD  through  the  pole  of  BC. 


106  SPHERICAL  TRIANGLES 

1.  R  .  COS.  B=cot.  AB  .  tan.  PB<90°. 

2.  Tan.  C  :  tan.  B  :  sin.  PB  :  sin.  PB<90' 
For  PB-|-PC  =  BC. 

3.  Sin.C  :  sin.  AB:  :sin.  B  :  sin  AC<90°. 

4.  Sin.  AC  :  sin.  B::sin.  BC  :  A  >90°. 
ForAB+BC>180°.-.A>BC(/. 


CASE   4. 

Given  two  angles  (=50°  and  62°  34'  6")   and    the   side 
FiM  34  between  them  (  =  110°)  to  find  the  other  parts. 

Project  the  Triangle  ;  making   CB  =  110°.    C=62°   34' 
6'',  and  B=50°  ;  also  draw  BPti  through  the  pole  of  CAe. 

1 .  R .  cos.  BC =cot.  C  .  cot  PBC  >  90° 
ThenPBC-ABC=PBA. 

2.  Sin.  BC.sin.  C=R  .  sin.  PB<90^ 

3.  R.  cos.  PBA=tan.  PB  .  cot.  BA<90«. 

4.  R.  cos.  BA=cot.  PBA  .  cot.  PAB<90. 
Then  100°-PAB  =  BAC>90. 

5.  R  .  COS.  BA=cos.  PB .  cos.  PA<90^. 
ThenPC-PA=AC<90. 


CASE  5. 

Given  the  three  sides  (=79*'  17'  14"  ,  110°  and  58*)  to 

find  the  angles. 

Fig.  33.       Project  the  Triangle  ;  making  AC=58°,  AB=79°    17' 

14"  and  BC  =  110°,  by  describing  small  circles  rg  and  pq, 

at  the  distance  of  79°  17'  14"  and  70°  from  A  and  e  as  poles 


SPHERICAL  TRIGONOMETRY.  197 

and  then  drawing  ABJ,  CBe  through  the  point  of  their  in- 
t€rsection. 

1.  Tan.  ^BC  :  tan,  iAC+AB::tan.    |AB  -  AC  :  tan. 
igp PC=  'D. 

Then  iBC+^iD=BP,  and  iBC  -  iD=PC. 
For  the  least  segment,  is  adjacent   to   the  least  side. 
(Sp.  Geom.  II  Prop.  XVII.) 

2.  The  other  parts  are  easily  found  as  before. 


CASE  6. 

Given  the  three  angles  (  =  121°54'  5G'\  SO^' and  62*  34' 
6'')  to  find  the  angles. 

The  supplements  of  these  angles,  are  the  sides  of  the 
supplemental  triangle,  the  angles  of  which  are  found,  as  in 
Case  5.  Then  the  supplements  of  those  angles  are  the 
sides  of  the  given  triangle. 

Spherical  Trigonometry  is  extensively  applicable  to  the 
lolution  of  questions  connected  with  Geography,  and  espe- 
cially in  Trigometrical  Surveying,  and  Geodesic  opera- 
tions.* 

Its  use  and  importance  in  astronomical  investigations,  is 
indicated  by  the  declaration  of  M.  de  La  Lande,  himself 
an  eminent  Astronomer,  ''  La  Trigonometrie  Spherique 
est  la  veritable  Science  de  VAstronome.^'' 

The  following  questions  will  illustrate  its  application  to 
each  of  these  sciences. 

1.  Given  the  Latitude  of  St.  Petersburg,  59^  66' N.  and 
its  Longitude,  27°  59'  30"  E.  from  Paris,  also  the  Lati- 
tude and  Longitude  of  Conception  in  South  America,  36° 
42'  53'  S.  arid  75°  W;  required  the  distance  of  the  two 
places,  as  measured  on  the  arc  of  a  great  circle.f 

In  the  triangle  NAB  two  sides  and  the  included  angle  P'-.^^'- 
are  given  ;  its  solution  therefore  is  according  to  Case  2,  of  '*'  * 
Oblique  angled  Triangles. 

*  Hutton'3  Math.  Vol.  II.  I  Cagr.oli  rilSO.^, 

26  -  ' 


198  SPHERICAL  TRIGONOMETRY. 

2.  Given  the  Longitude  and  Latitude  of  two  stars,  and 
the  distance  of  a  third  star  from  each  of  them,  required  the 
Longitude  and  Latitude  of  the  third  Star. 

I'V  4.  Let  EL  represent  the  Ecliptic  *PF,  PH  and  PL  arcs  of 
great  circles  perpendicular  to  it.  A  and  B  the  two  stars 
whose  places  are  given,  and  C  the  place  ofthe  third  star. 

Then  in  the  Triangle  APB,  two  sides  and  the  included 
angle  are  given ;  therefore  the  remaining  side  and  angles 
may  be  found. 

Then  in  the  triangle  ABC,  the  three  sides  are  given, 
whence  the  angles  may  be  found. 

Then  in  the  Triangle  PA(%  two  sides  (PA  and  AC)  are 
given,  and  the  angle  included  by  them.  Therefore  the 
remaining  side  PC,  (which  is  the  complement  of  the 
Latitude  of  C)  and  APC,  (the  difference  of  Longitude 
belwean  A  and  C)  may  be  found. 

If  the  Declination  and  Right  Ascension  of  each  of  the 
stars  A  and  B  were  given,  then  EL  would  represent  the 
Equator;  and  the  solution  would  give  the  Declination  dm^ 
Right  Ascension  ofthe  third  Star. 

*  The  Longitude  of  aheaveuly  body  is  measured  by  an  arc  of  the  Eclip* 
tic  and  its  Latitude  is  its  disianct  from  the  same  circle. 


NOTES  TO  CONIC  SECTIONS. 


Definitions. 

Conic  Sections  arc  treated  differently  by  different  au- 
thors ;  some  defining  them  by  the  sections  of  a  cone,  and 
deriving  their  properties  directly  from  the  intersection  ot* 
the  cone  and  plane,  while  others  define  them  as  figures  de- 
scribed in  a  plane,  and  derive  their  properties  from  their 
mechanical  description.  Each  method  has  its  advantages. 
If  the  latter,  in  some  cases,  renders  the  demonstrations 
more  simple  and  easy;  the  former,  which  is  adopted  in  this 
treatise,  has  the  important  advantage,  of  deriving  the  fig- 
ures from  solids  whose  properties  have  been  demonstrated, 
and  by  operations  which  have  become  familiar  in  the  Ele- 
ments of  Geometry. 

"  Thus  far,''  says  Newton,  •'  I  think  1  have  expounded 
the  construction  of  solid  Problems  by  operations  whose 
manual  practice  is  most  simple  and  expeditious.  So  the 
Ancients,  after  they  had  obtained  a  method  of  solving 
these  Problems  by  a  composition  of  solid  places,  thinking 
the  Constructions  by  the  Conick  Sections  as  useless,  by 
reason  of  the  Difficulty  of  describing  them,  sought  easier 
Constructions  by  the  Conchoid,  Cissoid,  the  Extension  of 
'threads,  and  by  any  Mechanic  Application  of  Figures.— 
"  If  the  Ancients  had  rather  construct  Problems  by  Fig- 
ures not  received  into  Geometry  at  that  Time,  how  much 
more  ought  these  Figures  now  to  be  preferred  which  are 
received  by  many  into  Geometry  as  well  as  the  Conick 
Sections. 

However  I  do  not  agree  to  this  new  sort  of  Geometri- 
cians who  receive  all  Figures  into  Geometry. -In  ray  judg- 
ment, no  Lines  ought  to  be  admitted  into  plain  Geometry 
besides  the  right  Line  and  the  Circle,  unless  some  Distinc- 


200  NOT£:s  TO  CONIC  SECTIONS. 

tion  of  Lines  might  be  first  invented,  by  which  a  circular 
Line  might  be  joined  with  a  right  Line,  and  separated 
from  all  the  rest.  But  truly  plain  Geometry  is  not  then 
to  be  augmented  by  the  number  of  Lines.  For  all  figures 
are  plain  that  are  admitted  into  plain  Geometry,  that  is, 
those  which  the  Geometers  postulate  to  be  described  in 
planoJ^"* — "  All  these  descriptions  of  the  Conicks  in  plano^ 
which  the  Moderns  are  so  fond  of,  are  foreign  to  Geome- 
try. Nevertheless,  the  Conick  Sections  ought  not  to  be 
flung  out  of  Geometry.  They  indeed  are  not  described 
Geometrically  mp/ffno,  but  are  generated  in  the  plane 
Superficies  of  a  Geometrical  Solid.  A  Cone  is  constituted 
geometrically,  and  cut  by  a  Geometrical  Plane.  Such  a 
segment  of  a  Cone  is  a  Geometrical  Figure,  and  has  the 
same  place  in  solid  Geometry,  as  the  Segment  of  a  Circle 
has  in  Plane,  and  for  this  reason  its  base,  which  they  call 
a  Conick  Section,  is  a  Geometrical  Figure.  Therefore  ft 
Conick  Section  hath  a  place  in  Geometry,  so  far  as  it  is  the 
Superficies  of  a  Geometrical  Solid  ;  <SiC.''  Universal  Arith, 
pp.  247-249. 

Def.  8.  — This  definition  was  suggested  by  a  communi- 
cation in  the  Journal  of  Science^  from  Prof  Davies  of  the 
Military  Academy  at  West  Point.  Jour.  ScL  VoL  VI  page 
280> 


ELLIPSE  PROPOSITION   XXVI. 
The  properties  referred  to,  are  the  following, 

PROPOSITION  A. 


If  any  line  in  the  Ellipse  pass  through  either  the  of  th« 
Foci,  and  a  tangent  be  drawn  through  one  of  its  extremities, 
a  line  drawn  from  the  centre,  parallel  to  the  line  passing 


i\OTES  TO  CONIC  SECTIONS.  201 

through  the  Focus  and  intercepted  by  the  tangent,  is  equal 
to  the  semi-transverse  axis. 
That  is,  C^=CA.  Fig.  26. 

Bor,  (IX  Cor.  1.^  CF  .  CD=CA2-CA  .  EF, 

CA.EF=CA2-CF.CD; 

but,  (VI;  CA»=CD  .  CT ;  ..CA  .  EF=CD|.  CT 


[-CD.CF=CD.TF, 
:TF. 


sim.   tri.   TC  :  C/:  :TF  :  FE:  :TF  .  CD  (=CA  .  FE) 

[FE  .  CD, 
::CA:CD, 
therefore,  Ct ,  CA=TC  .  CD^CA^. 
Ct=CA. 

Q.  E.  D, 

CoR.  1.— Hence  TF  .  CD=CA  .  FE.  3rf  step  ofdem. 

Cor.  2. — AB  :  kz  :  EH  are  continued  proportionals. 

For,  draw  the  diameter  LCIK,  bisecting  EH,  and  draw 
Ed  parallel  to  it. 

Then,  (XXVI.)  Cd:  Ck  :  C^,are  continued  proportionals, 
That  is  IE:CA::CA,      '' 

HE:zA::BA,      " 
and  zA^^HE.BA. 


PROPOSITION  B. 


The  rectangle  of  the  segments  of  any  line  in  the  Ellipse, 
passing  through  the  Focus,  is  equal  to  the  rectangle  of  one 
fcurth  of  the  Parameter,  into  the  same  line. 

That  is,  EF  .  FH  =  iP  .  EH  Fig.  §6 


502  NOTES  TO  CONIC  SECTIONS. 

For,  (XXIV,)  Ck^  :  CA« :  :EF  .  FH  :  AF  .FB  ; 
But  (VII.  Cor.  3  )  AF    FB  =  (Ca^=)  CA  .  iP, 
Therefore,  C^^rEF.FH.-rCA^'  :  CA  .  iP," 
But,  CA  :  \P\  :CA  .  EI  :  iP  .  EI, 
And,CA2  :CA.  iP:  :CA  .  EI  :  i  P  .  EI. 
.-.     Ck^  :EF.FH::CA.EI:iP.  EI. 
But,  fXXVII.  Cor.  2  )  CA:-^  =EI.  CA.   EF  .  FH=iP  . 

[EI,  or  lEH  .  iP, 
...     EF.FH=iP.EH. 

q.  E.  d: 

CoR.  1. — As  the  proposition  is  applicable  to  all  lines 
passing  through  the  Focus, 

AF  .  FB  =  iP  .  AB,  as  demonstrated.     Prop.  VII. 

AndEF.  FH  =  iP.EH, 

Therefore,  AB  :  EH:  :AF  .  FB  :  EF  .  FH.  (See  Pa- 
rab.  XIV.  Cor.  7.) 

Cor.  2. — When  EH,  is  perpendicular  to  the  Transverse 
Axis,  it  is  a  double  ordinate  to  it,  and  since  the  rectangle  of 
the  abscisses  at  the  Focus  equals  the  square  of  the  semi- 
conjugate,  and  EH  then  equals  the  parameter,  and  the 
rectangle  EF  .  FH  is  the  same  as  the  square  of  EF,  the 
proposition  becomes 

As  the  Transverse  Axis 

Is  to  the  Parameter, 

So  is  the  square  of  the  Semi-Conjugate 

To  the  square  of  the  Focal-Ordinate. 


PROPOSITION  C. 

The  rectangle  of  the  two  lines   drawn  from  the  Foci,  to 
any  point  in  the  curve,  is  equal  to  the  square  of  half  the  di- 
ameter conjugate  to  that  which  passes  through  the  point. 
Fig.  12.      ThatisFE./E=CpS 


CURVATURE  OF  CONIC  SECTIONS.  2()3 

Let  ER  be  drawn  perpendicular  on  eh, 
Then,  sim.  triang.  FE :  FP:  :EI  :  ER, 
And      •«         ''      yE  :/)?::EI:ER, 
Therefore  FE  .  /"E  :  FP  .^:  EP  :  ER^ 

But,  (IX.  Cor.  4.)  FP  - j5?=Ca2  ;  and  EP  =€?«=( VII) 

Therefore,  FE  ./E  :  Ca^ ;  iCA^  :  ER^ ; 

But,  (XVI.)  Ce  .  ER=AC  .  Ca, 
.:.  Ce:Ca::CA  :  ER, 
And,  Ce^  iCa^iiCA^  :  ER% 
Therefore,  FE./E=Ce^ 

Q.  £.  Z>. 


CURVATURE  OF  CONIC  SECTIONS. 


The  application  of  Conic  Sections  to  Physical  Astrono- 
my, which  is  one  of  their  most  important  applications,  re- 
quires an  acquaintance  with  their  Curvatures,  and  espe- 
cially with  the  method  of  reasoning  employed  in  treating 
of  this  part  of  Conic  Sections.  {See  Kewtons  Principia 
Lib*  I  Sect.  1.  Cavalh^s  Philosophy,  Introduction,  Lem- 
mas, Enfield'' s  Phil,  Central  Forces,  Lem.)  As  the  doc- 
trine of  the  Curvature  of  Conic  Sections,  gives  the  stu- 
dent new  and  interesting  views  of  their  general  properties, 
and  is  attended  with  no  peculiar  difficulties,  a  few  elemen- 
tary propositions  are  introduced  in  this  part  of  the  course* 


NOTES  TO  SPHERICAL  TRIGONOMETRY, 


PROPOSITION  V. 

•■•The  angles  at  the  base  of  an  isosceles  spherical  triangle 
are  Symttrkal  magnitudes,  not  admitting  of  being  laid  on 
one  another,  nor  of  coinciding,  notwithstanding  their 
equality.  It  might  be  considered  as  a  sufficient  proof  that 
they  are  equal,  to  observe  that  they  are  each  determined 
to  be  of  a  certain  magnitude  rather  than  any  other,  by 
conditions  which  are  precisely  the  same,  so  that  there  is 
no  reason  why  one  of  them  should  be  greater  than  anoth- 
er. For  the  sake  of  those  to  whom  this  reasoning  may  not 
prove  satisfactory,  the  demonstration  below  is  given, 
which  is  strictly  geometrical." 

Playfair^s  Sph,  Ti-ig. 


.  „y.       Let  ABC  be  a  spherical   triangle,  havmg  the  side  AB 
Fig.  1.  equal  to  the  side  AC  ;  the  spherical  angles  ABC  and  ACB 
are  equal. 

Let  D  be  the  centre  of  the  sphere  ;  join  DB,  DC,  DA, 
and  from  A  on  the  straight  lines  DB,  DC,  draw  the  per- 
pendiculars AE,  AF  ;  and  from  the  points  E  and  F  draw 
in  the  plane  DBC  the  straight  lines  EG,  FG  perpendicu- 
lar to  DB  and  DC,  meeting  one  another  in  G  ;  Join  AG. 
Because  DE  is  at  right  angles  to  each  of  the  straight 
lines  AE.  EG,  it  is  at  right-angles  to  the  plane  AEG,  which 


NOTES  TO  SPHERICAL  TRIGONOMETRY.  40" 

passes  through  AE,  EG  (4.2.  Sup.)  ;  and  therefore,  every 
plane  that  passes  throu^  DE  is  at  right-angles  to  the  plane 
AEG  (17.  2.  Sup.)  ;  wherefore,  the  plane  DBC  is  at  right- 
angles  to  the  plane  AEG.  For  the  same  reason,  the  plane 
DBC  is  at  right-angles  to  the  plane  AFG,  and  therefore 
AG,  the  common  section  of  the  planes  AFG,  AEG  is  at 
right-angles  (18.  2.  Sup.)  to  the  plane  DBC,  and  the  an- 
gles AGE,  AGF  are  conseqnently  right-angles 

But  since  the  arch  AB  is  equal  to  the  arch  AC,  the  an- 
gle ADB  is  equal  to  the  angle  ADC.  Therefore  the  tri- 
angles ADE,  ADF,  have  the  angles  EDA,  FDA  equal,  as 
also  the  angles  AED,  AFD,  which  are  right-angles  ;  and 
they  have  the  side  AD  common,  therefore  the  other  sides 
are  equal,  viz.  AE  to  AF,  (26.  1.),  and  DE  to  DF.  Again, 
because  the  angles  AGE,  AGF  are  right-angles,  tho 
squares  on  AG  and  GE  are  equal  to  the  square  of  AE  ; 
and  the  squares  of  AG  and  GF  to  the  square  of  AF.  But 
the  squares  of  AE  and  AF  are  equal,  therefore  the  squares 
of  AG  and  GE  are  equal  to  the  squares  of  AG  and  GF, 
and  taking  away  the  common  square  of  AG,  the  remain- 
ing squares  of  GE  and  GF  are  equal,  and  GE  is  therefore 
equal  to  GF.  Wherefore,  in  the  triangles  AFG,  AEG, 
the  side  GF  is  equal  to  the  side  GE,  and  AF  has  been 
proved  to  be  equal  to  AE,  and  the  base  AG  is  common, 
therefore,  the  angle  AFG  is  equal  to  the  angle  AEG  (8. 
1.).  But  the  angle  AFG  is  the  angle  which  the  plane  ADC 
makes  with  the  plane  DBC  (4.  def.  2.  Sup.)  because  FA 
and  FG,  which  are  drawn  in  these  planes,  are  at  right-an- 
gles to  DF,  the  common  section  of  the  planes.  The  an- 
gle AFG  (3.  def)  is  therefore  equal  to  the  spherical  angle 
ACB  ;  and,  for  the  same  reason,  the  angle  AEG  is  equal 
to  the  spherical  angle  ABC.  But  the  angles  AFG,  AEG 
are  equal.  Therefore  the  spherical  angles  ACB  ABC 
are  also  equal. 

Q.  E.  D. 

The    converse  of  this  proposition  is    thus  demonstrated  by 
PI  ay  fair,  • 

Let  ABC  be  a  spherical  triangle  having  the  angles  ABC 
ACB  equal  to  one  another  ;  the  sides  AC  and  AB  are  al- 
so equal. 

27 


408  NOTES  TO  SPHERICAL  TRIGONOMETRY. 

Let  D  be  the  centre  of  the  sphere  ;  join  DB,  DA,  DC, 
and  from  A  on  the  straight  hnes  DB,  DC,  draw  the  per- 
pendiculars AE,  AF  ;  and  from  the  points  E  and  F,  draw 
in  the  plane  DBC  the  straight  lines  EG,  FG  perpendicu- 
lar to  DB  and  DC,  meeting  one  another  in  G  ;  join  AG. 

Then,  it  may  be  proved,  as  was  done  in  the  last  propo- 
sition, that  AG  is  at  right  angles  to  the  plane  BCD,  and 
that  therefore  the  angles  AGF,  AGE  are  right-angles, 
and  also  that  the  angles  AFG,  AEG  are  equal  to  the  an- 
gles which  the  planes  DAC,  DAB  make  with  the  plane 
DBC.  But  because  the  spherical  angles  ACB,  ABC  are 
equal,  the  angles  which  the  planes  DAC,  DAB  make  with 
the  plane  DBC  are  equal,  (3.  def.)  and  therefore  the  an- 
gles AFG,  AEG  are  also  equal.  The  triangles  AGE, 
AGF  have  therefore  two  angles  of  the  one  equal  to  the  two 
angles  of  the  other,  and  they  have  also  the  side  AG  com 
mon,  wherefore  they  are  equal,  and  the  side  AF  is  equal 
to  the  side  AE. 

Again,  because  the  triangles  ADF,  ADE  are  right-an- 
gled at  F  and  E,  the  squares  of  DF  and  FA  are  equal  to  the 
square  of  DA,  that  is,  to  the  squares  of  DE  and  DA  ,  now,  the 
square  of  AF  is  equal  to  the  square  of  AE,  therefore  the  * 
square  of  DF  is  equal  to  the  square  of  DE,  and  the  side 
DF  to  the  side  DE.  Therefore,  in  the  triangles  DAF, 
DAE,  because  DF  is  equal  to  DE,  and  DA  common,  and 
also  AF  equal  to  AE,  the  angle  ADF  is  equal  to  the  an- 
gle ADE  ;  therefore  also  the  arches  AC  and  AB,  which 
are  the  measures  of  the  angles  ADF  and  ADE,  are  equal 
to  one  another  ;  and  the  triangle  ABC  is  isosceles. 

Q.  £.  D. 


ERRATA. 


The  following  errors  of  the  press  should  be  corrected  with 
a  pen  or  pencil,  before  reading  the  book. 

Page  10  line  4  fr.  top,  for  XII  Prop.  2,  read  Sup.  I,  Prop. 

47  Prop.  IX       „  /E4-/E  „  FE+/E, 

51  line  13  fr.  top  .,  iFG  „  ^/G, 

52  „      7  „     „     „  PAC „  P  .  AC^ 

„    „    21  „     „     „  CF  :  CF  +  FT  „  CF  .  CF+FT, 

54  Prop:  XIV      „  De^  and  D  e^     „  DE^  and  D  E^ 

55  line  5  fr.  hot.    „  Ca,  „  CA, 
..     "     2     „     „     ,,  Ggr  „C^, 
57    „  10     „     „     „  GMT,            „  CMT, 
178,,  2fr.  top,       „  iBD-AD,  „  iBD-AD, 
185  „  2  „     „         „  lAC+AB  and  j  AC  -ABread 

lAC+ABandiAC— AB. 
204  for  Prop.  V,  read  Prop.  IV. 


ERRATA. 


[Of  the  following  Errata,  some  were  obviously  owing  to  the  press,  and  many  to  inaccura- 
cies of  tbe  manuscript,  which  had  been  transcribeil,  and  was  written  originally  in  haste,  and 
not  very  legibly  A  considerable  number  of  them  are  unimportant,  but  it  was  thought 
zulviseable  to  make  tbe  list  as  complete  as  possible.] 


12 

margin,    - 

-     .    for    -     -    66    -     - 

read 

-    -    6.8. 

15 

24th  line  from 

top     -     "     -     -   Fg3        - 

(( 

-          /«2. 

17 

10 

{( 

«k 

bottom,      insert    that  o/j  after  equal  to.  ~ 

20 

9 

(( 

(( 

top,      for     AF 

read 

AF.2 

_ 

12 

« 

h 

do.        '*    2AF 

it 

-2AF. 

23 

last  line, 

- 

«     (VII) 

(( 

(VI). 

27 

5 

t4 

from  top,        "     circle  - 

{( 

curve. 

_ 

27 

(( 

" 

do.         "     II  Cor.      - 

n 

(XI  Cor.  4.) 

29 

11 

(1 

(( 

bottom,  "    (II  Cor.  1) 

(( 

(XI  Cor.  4.) 

- 

2 

n 

do.      "    (Viii;       - 

« 

(XI; 

30 

6 

u 

top,         "     2CIVI:TP 

u 

2CM.TP. 

34 

- 

margin,  «     Fig.  6.  Fl.  VII 

(( 

PI.  VIII.  Fig.  14. 

35 

1&2 

a 

bottom,  "    ad.ab 

<; 

ad.db. 

38 

14 

n 

top,         "    eLkl 

<{ 

ik.kt. 

46 

18 

« 

do.         «     (III. 

« 

(11. 

_ 

22 

i( 

do.         '^      FF2 

t( 

F/3. 

47 
49 

16 

do.         *«   /E±/-E      - 
do.          "    TF 

tt 

FE+FE. 
T/. 

50 

12 

(( 

do.         "    OEP 

ii 

OE/. 

51 

8 

(( 

do.           "    fep, 

ti 

/Ef 

- 

13 

(( 

do.          ♦*     iFG 

u 

i/£. 

. 

4 

u 

bottom,  "    semi-diameter 

a 

diameter. 

- 

- 

n 

''   /PK           . 

i( 

»PK. 

52 

7 
21 

<( 
u 

top,          «     PAC 

do.          "    CF:CF-I-FT 

(4 

P.AC- 

— 

CF.CF-I-FT. 

54 

lO&ll  « 

n 

do.          "    De2D'c2 

(( 

DE2  D'Ea . 

55 

- 

- 

margin,  "     Fig.  14.     - 

i< 

Fig.  13. 

- 

5 

« 

u 

bottom,  "    Ca 

u 

CA. 

- 

2 

ti 

u 

do.      «     G^ 

i( 

Cg- 

56 

5 

it 

(( 

top,         "     De2 

<( 

de2  . 

57 

10 

(t 

ti 

bottom,  **    GMT 

« 

CMT. 

60 

8 

(( 

(i 

do.       «    + 

(C 

64 

- 

- 

top,         «    HK 

(( 

kK. 

- 

2&3 

<( 

(( 

do.         «'    DT 

(( 

BT. 

- 

- 

- 

margm, '«    Fig2  4.    - 

'4 

Fig.  25. 

- 

6 

U 

« 

top,         «    -f. 

£( 

g.    •««*. 

75 

- 

- 

do.         "    CL 

t( 

CA. 

- 

3 

a 

n 

do.         "    CD 

<( 

CT. 

77 

- 

- 

margin,  "    Fig.  10.     - 

u 

Fig.  9. 

78 

6 
15 

top,         «    BHF 
do.          "    (III. 

it 

BH/. 

2  ERKATAi 

Page    78     I7th  line  from  top,        for  CA2  -  CP2      read  CF8  -  CAa  , 

-  2        <t      ««    bottom,  insert //letr  «7wares  after  are  c^wa/. 

81  13        <'      "    top,        for  FE/         -      read  PEP. 

82  -         -         bottom,  "     semi- diameter      "  diameter. 

-  -         -  do.        "   /PK  -         '«  p?K, 

87  5        '»      ♦*    top,         "     De2  -  «        rfe^  . 

88  15        "      **     do.  "    CHN        -         "  CHA. 

91  5        "      *♦     do.         *♦     CA2  -         «  Ca2 

-  11        "      '<     do.         '•    heeh         -         *<        /i«.eAr. 

margin,  ♦'    Fig.  14.     -         "         Fig.  16. 

92  4        "      «    top,         '*    AEK         -         *<        CEK. 

96  3,5  &6  "      "    bottom,  invert  the  terms  of  the  following  praposi- 

tions— iviz : 

MT  is  less  than  MA 

CM  is  less  than  MA 

CM  a  is  less  than  MA2 

99      6        <«      •<    top,         "    diameter  read      radius. 

-  11&22*'      "    do.  «*    CB2  -         *'        Cr2  .CF. 

-  19        **      «'    do.     insert  .-.Cra. FA       after      CV2  . 

-  4        «      <t    bottom,  note,   omit    varies  as  FC  a . 

102      9        K      a    do.     insert  the  cube  of      after       inversely  as. 

-  1,2&3  "      "    do.    insert  inversely^        after       varixs^   and 

for    HC         -        read      He. 
110    13        "      »'    top,        "     BA  -  «         CA. 

112  9        «       "    do.         "    CA  -  ♦'         Ca. 

113  -         -         do.  *'    FNC       -  **         FVC, 

114  13        «      "    bottom,  "    NP  -  «         NF. 

-  11        •«      «    do.         «'    NEA       -  «*         NFA. 

7        «      ♦'    do.  «    ER  -         »         ET. 

117      9        "      "    do.  '<    CA:Ca  «         Co:Ca'. 

119      6       *•      "    bottom,  «*    LIR        -  **         LIK. 

"      *»    do.  »'    PiKh        -  »         AJfeH. 

2        **      '*    do.  "    EKG       -  «'         EKff. 

135      4        «*      «    do.  "    PEp        -  «'         PF/7. 

139  20        '<      "    do.  *♦     ABC        -  «  ABD. 

140  3        "      '<    do.  ««    AEC       -  '<         EAC. 

-  4        **      "    do.  «    AC  -  '<         EC. 

-  16        ft      "    do.  "    C,  B&A  "  C,A&B. 

-  18        «      '«    do.         *<    DF,  FE  &  ED  «         AB,  BC  &  CA. 

143  23        «      «    do.  "     <^  .  «  V, 

-  24        ft      ft    do.  ft    CEA        -  «         CED. 

144  -        -         margin,    dele  Fig.  12. 

148  18        "      "    top,       for   FG  «         EG. 

149  3        ft      "    do.  **    Cor.  7      -  '•         Cop.  U. 

-  8        «      »*    do.  *<    BDC       -  «         ADC. 
153      9        *»      "    bottom,  ft     \if           -          «         Vif. 
155     14        "      ft    do.          "    HIi         -          «         Mi. 
157        -         -          margin,  «    PI.  XIV.             *«         PI.  XVI. 

160    13        ft       "    top,        "    axis  of  the  eye^  «     plane  of  prx>jeet%on. 

166    14        **      "    do.      insert  and  H  /o  P    after  reduce  P^  to  G. 

-  15        «      «    do.        for    Y,  P  &  G        read       P,  Y  &  p. 
166      9        '«      "    bottom,  ft    TV  -  ft         TH, 

and  after  VY  insert    through  X  w/ierc  EH  tnter- 

sects  the  great  circle,  whose  pole  is  E.  (Cor.  3.) 

171  10        ft      ft    top,        "     AB,  AD,  AC,  read      DB.DA,  &  DC. 

172  5        ft      «    do.  "ABC        -  ft         tang.  ABC 
174      5        '•      ft    bottom,**    Cos.  A.    -         **         Cos.  CA 


ERRATA. 

176 

5th  line  from  bottom, 

for 

tang.  BC 

read 

tang.  DC. 

178 

2 

n 

(I 

top. 

(( 

iBD— AD 

t( 

iBD— AD 

177 

11 

il 

(( 

do 

(( 

+ 

u 

^ 

15 

i( 

l( 

do. 

(( 

I 

T 

(•' 

i 

^ 

48 

(( 

t( 

do. 

(4 

BB 

(i 

BC. 

^ 

4 

(( 

u 

bottom. 

rad 

n 

radi  . 

_ 

5 

i( 

u 

do. 

i» 

cot 

a 

xoU 

181 

20 

u 

(( 

top, 

u 

B 

u 

C. 

185 

2 

{( 

(( 

do. 

(( 

iAC+AB  &  iAC— 

AB, 

read 

iACtAB  &  iAC- 

AB. 

189 

7 

(( 

(( 

do. 

for 

AP 

read 

Ap. 

11 

(( 

(( 

do. 

(( 

y\C 

(( 

^A/. 

„ 

15 

« 

<c 

do. 

a 

BC 

u 

Cos  BC. 

191 

13 

(( 

« 

do. 

il 

fad 

PI.  XVI      - 

u 

/Arf. 
PI.  XV. 

195 

6 

( 

(( 

do. 

a 

li 

196 

2 

u 

(( 

do. 

it 

PB 

(( 

PC. 

204 

_ 

- 

top, 

u 

TRIGONOMBTRY    "        GEOMETRY, 

and 

(I 

PROP.  V 

(( 

PROP.  IV, 

The  assertion  at  the  top  of  page  144  is  obviously  false — the  converse  of  it 
is  true.  Also,  in  the  corollary,  at  the  bottom  of  the  next  page,  the  assertion 
implied  in  the  last  clause  of  each  sentence,  is  not  necessarili/ trve  but  of  on£ 
aide,  and  one  angle.  M.  R.  D. 


1'LA.TV:     I. 


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PLATE      XI, 


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PLATE     XiV 


PLATE     XIV 


PLATE  XV. 


A.BoclUtlc 


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UNIVERSITY  OF  CALIFORNIA  LIBRARY 

Los  Angeles 
This  book  is  DUE  on  the  last  date  stamped  below. 


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MAR  17  1972 


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